# Differentiation, finding stationary points + determining maxima or minima

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I have recently been introduced to calculus and differentiation which I have really developed a passion and admiration toward. That being said, I would not claim to have the firmest footing in terms of my comprehension of the topic. Therefore, while revising I ran across some practise questions which I have solved and I wonder if anyone would look over my calculations to firstly check whether they are correct and secondly to review whether I should have used a better method or likewise. I would really appreciate any feedback ✌️

a) The question states; find the stationary points of the curve y=2x^3-3x^2-12x-7 and state whether they are maximum or minimum points.

So stationary points are where the gradient is zero, i.e. where dy/dx = 0.

To begin, differentiate y=2x^3-3x^2-12x-7:

dy/dx = 6x^2-6x-12

Factor out the common denominator:

6(x^2-x-2)=0

Factor x^2-x-2:

(x-2)(x+1)=0

Therefore, x=2 or x =-1

When x= 2, y=2(2)^3-3(2)^2-12(2)-7=-27

When x= -1, y= 2(-1)^3-3(-1)^2-12(-1)-7=0

The turning points are therefore (-1,0) and (2,-27)

To determine whether these are maxima or minima one should examine the pattern of the gradient dy/dx.

The gradient either side of x=-1:

x=-1.1, dy/dx=6(-1.1)^2-6(-1.1)-12= 1.86=+ve

x=-1, dy/dx=6(-1)^2-6(-1)-12=0 (zero)

x=-0.9, dy/dx=6(-0.9)^2-6(-0.9)-12=-1.74=-ve

So the gradient shows a pattern of +ve, zero, -ve, which shows a maximum point.

The gradient either side of x=2:

x=1.9, dy/dx=6(1.9)^2-6(1.9)-12= -1.74=-ve

x=2, dy/dx=6(2)^2-6(2)-12=0 (zero)

x=2.1, dy/dx=6(2.1)^2-6(2.1)-12=1.86=+ve

The gradient the gradient shows a pattern of -ve, zero, +ve, which shows a minimum point.

Therefore, there is a minimum point at (2,-27) and a maximum point at (-1,0)

b) Differentiate the following with respect to x:

i. (3x^2+2)^2

ii. 2x^2(3-4x)

This is where I am having the most difficulty. I have been taught to differentiate using y = x^n, dy/dx = nx^n-1 and also from first principles but I do not know whether either method is suitable here? I have, in my research, stumbled upon implicit differentiation and the chain rule but I do not really understand these methods, however, would they be more suitable here?

Do my answers appear correct? Thank you in advance I really appreciate any and all responses 😁

I have recently been introduced to calculus and differentiation which I have really developed a passion and admiration toward. That being said, I would not claim to have the firmest footing in terms of my comprehension of the topic. Therefore, while revising I ran across some practise questions which I have solved and I wonder if anyone would look over my calculations to firstly check whether they are correct and secondly to review whether I should have used a better method or likewise. I would really appreciate any feedback ✌️

a) The question states; find the stationary points of the curve y=2x^3-3x^2-12x-7 and state whether they are maximum or minimum points.

So stationary points are where the gradient is zero, i.e. where dy/dx = 0.

To begin, differentiate y=2x^3-3x^2-12x-7:

dy/dx = 6x^2-6x-12

Factor out the common denominator:

6(x^2-x-2)=0

Factor x^2-x-2:

(x-2)(x+1)=0

Therefore, x=2 or x =-1

When x= 2, y=2(2)^3-3(2)^2-12(2)-7=-27

When x= -1, y= 2(-1)^3-3(-1)^2-12(-1)-7=0

The turning points are therefore (-1,0) and (2,-27)

To determine whether these are maxima or minima one should examine the pattern of the gradient dy/dx.

The gradient either side of x=-1:

x=-1.1, dy/dx=6(-1.1)^2-6(-1.1)-12= 1.86=+ve

x=-1, dy/dx=6(-1)^2-6(-1)-12=0 (zero)

x=-0.9, dy/dx=6(-0.9)^2-6(-0.9)-12=-1.74=-ve

So the gradient shows a pattern of +ve, zero, -ve, which shows a maximum point.

The gradient either side of x=2:

x=1.9, dy/dx=6(1.9)^2-6(1.9)-12= -1.74=-ve

x=2, dy/dx=6(2)^2-6(2)-12=0 (zero)

x=2.1, dy/dx=6(2.1)^2-6(2.1)-12=1.86=+ve

The gradient the gradient shows a pattern of -ve, zero, +ve, which shows a minimum point.

Therefore, there is a minimum point at (2,-27) and a maximum point at (-1,0)

b) Differentiate the following with respect to x:

i. (3x^2+2)^2

ii. 2x^2(3-4x)

This is where I am having the most difficulty. I have been taught to differentiate using y = x^n, dy/dx = nx^n-1 and also from first principles but I do not know whether either method is suitable here? I have, in my research, stumbled upon implicit differentiation and the chain rule but I do not really understand these methods, however, would they be more suitable here?

Do my answers appear correct? Thank you in advance I really appreciate any and all responses 😁

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a) The question states; find the stationary points of the curve y=2x^3-3x^2-12x-7 and state whether they are maximum or minimum points.

So stationary points are where the gradient is zero, i.e. where dy/dx = 0.

To begin, differentiate y=2x^3-3x^2-12x-7:

dy/dx = 6x^2-6x-12

Factor out the common denominator:

6(x^2-x-2)=0

Factor x^2-x-2:

(x-2)(x+1)=0

Therefore, x=2 or x =-1

When x= 2, y=2(2)^3-3(2)^2-12(2)-7=-27

When x= -1, y= 2(-1)^3-3(-1)^2-12(-1)-7=0

The turning points are therefore (-1,0) and (2,-27)

To determine whether these are maxima or minima one should examine the pattern of the gradient dy/dx.

The gradient either side of x=-1:

x=-1.1, dy/dx=6(-1.1)^2-6(-1.1)-12= 1.86=+ve

x=-1, dy/dx=6(-1)^2-6(-1)-12=0 (zero)

x=-0.9, dy/dx=6(-0.9)^2-6(-0.9)-12=-1.74=-ve

So the gradient shows a pattern of +ve, zero, -ve, which shows a maximum point.

The gradient either side of x=2:

x=1.9, dy/dx=6(1.9)^2-6(1.9)-12= -1.74=-ve

x=2, dy/dx=6(2)^2-6(2)-12=0 (zero)

x=2.1, dy/dx=6(2.1)^2-6(2.1)-12=1.86=+ve

The gradient the gradient shows a pattern of -ve, zero, +ve, which shows a minimum point.

Therefore, there is a minimum point at (2,-27) and a maximum point at (-1,0)

**LukeWatson4590**)a) The question states; find the stationary points of the curve y=2x^3-3x^2-12x-7 and state whether they are maximum or minimum points.

So stationary points are where the gradient is zero, i.e. where dy/dx = 0.

To begin, differentiate y=2x^3-3x^2-12x-7:

dy/dx = 6x^2-6x-12

Factor out the common denominator:

6(x^2-x-2)=0

Factor x^2-x-2:

(x-2)(x+1)=0

Therefore, x=2 or x =-1

When x= 2, y=2(2)^3-3(2)^2-12(2)-7=-27

When x= -1, y= 2(-1)^3-3(-1)^2-12(-1)-7=0

The turning points are therefore (-1,0) and (2,-27)

To determine whether these are maxima or minima one should examine the pattern of the gradient dy/dx.

The gradient either side of x=-1:

x=-1.1, dy/dx=6(-1.1)^2-6(-1.1)-12= 1.86=+ve

x=-1, dy/dx=6(-1)^2-6(-1)-12=0 (zero)

x=-0.9, dy/dx=6(-0.9)^2-6(-0.9)-12=-1.74=-ve

So the gradient shows a pattern of +ve, zero, -ve, which shows a maximum point.

The gradient either side of x=2:

x=1.9, dy/dx=6(1.9)^2-6(1.9)-12= -1.74=-ve

x=2, dy/dx=6(2)^2-6(2)-12=0 (zero)

x=2.1, dy/dx=6(2.1)^2-6(2.1)-12=1.86=+ve

The gradient the gradient shows a pattern of -ve, zero, +ve, which shows a minimum point.

Therefore, there is a minimum point at (2,-27) and a maximum point at (-1,0)

Secondly, you can speed up the process of determining the nature of the stationary points. What you have at the moment is a good understanding, and now you can extend this slightly more and employ the second derivative to your advantage.

For example, the point x=-1. You have observed that as we go through this point, the gradient is decreasing because it goes from being +ve to being 0 to being -ve. But think of the gradient as a function. To show that this gradient is decreasing at this point, you just need to show that the function is decreasing at this point.

And this should ring an alarm bell or two in your head if you have heard this phrasing before, because to show a function is decreasing at a point, we just need to show that its derivative is negative. Hence, we just need to show that the derivative of the gradient (i.e. the second derivative of the function y) is negative at x=-1. Hence the point is a maximum.

b) Differentiate the following with respect to x:

i. (3x^2+2)^2

ii. 2x^2(3-4x)

This is where I am having the most difficulty. I have been taught to differentiate using y = x^n, dy/dx = nx^n-1 and also from first principles but I do not know whether either method is suitable here? I have, in my research, stumbled upon implicit differentiation and the chain rule but I do not really understand these methods, however, would they be more suitable here?

i. (3x^2+2)^2

ii. 2x^2(3-4x)

This is where I am having the most difficulty. I have been taught to differentiate using y = x^n, dy/dx = nx^n-1 and also from first principles but I do not know whether either method is suitable here? I have, in my research, stumbled upon implicit differentiation and the chain rule but I do not really understand these methods, however, would they be more suitable here?

Chain rule can be applied here but I fear that if you have not formally been taught it then there is no use trying it. Implicit differentiation does not apply here.

The simplest thing to do here is to expand the brackets fully. And then differentiate term by term using the basic formula you have stated.

Last edited by RDKGames; 1 year ago

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(Original post by

This is good and clear working out. Be careful when you say "factor our the common denominator" because it's not what you're doing. I think you just phrased this incorrectly but meant the correct approach which is to take out a common factor of 6.

Secondly, you can speed up the process of determining the nature of the stationary points. What you have at the moment is a good understanding, and now you can extend this slightly more and employ the second derivative to your advantage.

For example, the point x=-1. You have observed that as we go through this point, the gradient is decreasing because it goes from being +ve to being 0 to being -ve. But think of the gradient as a function. To show that this gradient is decreasing at this point, you just need to show that the function is decreasing at this point.

And this should ring an alarm bell or two in your head if you have heard this phrasing before, because to show a function is decreasing at a point, we just need to show that its derivative is negative. Hence, we just need to show that the derivative of the gradient (i.e. the second derivative of the function y) is negative at x=-1. Hence the point is a maximum.

First principles are the formal way to tackle derivatives, however it's best to get good with algebra first before employing that on such a question.

Chain rule can be applied here but I fear that if you have not formally been taught it then there is no use trying it. Implicit differentiation does not apply here.

The simplest thing to do here is to expand the brackets fully. And then differentiate term by term using the basic formula you have stated.

**RDKGames**)This is good and clear working out. Be careful when you say "factor our the common denominator" because it's not what you're doing. I think you just phrased this incorrectly but meant the correct approach which is to take out a common factor of 6.

Secondly, you can speed up the process of determining the nature of the stationary points. What you have at the moment is a good understanding, and now you can extend this slightly more and employ the second derivative to your advantage.

For example, the point x=-1. You have observed that as we go through this point, the gradient is decreasing because it goes from being +ve to being 0 to being -ve. But think of the gradient as a function. To show that this gradient is decreasing at this point, you just need to show that the function is decreasing at this point.

And this should ring an alarm bell or two in your head if you have heard this phrasing before, because to show a function is decreasing at a point, we just need to show that its derivative is negative. Hence, we just need to show that the derivative of the gradient (i.e. the second derivative of the function y) is negative at x=-1. Hence the point is a maximum.

First principles are the formal way to tackle derivatives, however it's best to get good with algebra first before employing that on such a question.

Chain rule can be applied here but I fear that if you have not formally been taught it then there is no use trying it. Implicit differentiation does not apply here.

The simplest thing to do here is to expand the brackets fully. And then differentiate term by term using the basic formula you have stated.

Excellent, I will evaluate the second derivative in future when determining whether a stationary point is a maximum or minimum.

Right so for part b) i. Expanding (3x^2+2)^2 = (3x^2+2)(3x^2+2)

9x^4+12x^2+4

dy/dx = 36x^3+24x

ii. Expanding 2x^2(3-4x):

6x^2-8x^3

dy/dx=12x-24x^2

Is this correct? Thank you very much for your response and your advice it has been tremendously helpful 😁

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(Original post by

Sorry, yes I was typing quickly and did not realise I had erroneously written 'the common denominator' and I did indeed mean the common factor, being 6.

Excellent, I will evaluate the second derivative in future when determining whether a stationary point is a maximum or minimum.

Right so for part b) i. Expanding (3x^2+2)^2 = (3x^2+2)(3x^2+2)

9x^4+12x^2+4

dy/dx = 36x^3+24x

ii. Expanding 2x^2(3-4x):

6x^2-8x^3

dy/dx=12x-24x^2

Is this correct? Thank you very much for your response and your advice it has been tremendously helpful 😁

**LukeWatson4590**)Sorry, yes I was typing quickly and did not realise I had erroneously written 'the common denominator' and I did indeed mean the common factor, being 6.

Excellent, I will evaluate the second derivative in future when determining whether a stationary point is a maximum or minimum.

Right so for part b) i. Expanding (3x^2+2)^2 = (3x^2+2)(3x^2+2)

9x^4+12x^2+4

dy/dx = 36x^3+24x

ii. Expanding 2x^2(3-4x):

6x^2-8x^3

dy/dx=12x-24x^2

Is this correct? Thank you very much for your response and your advice it has been tremendously helpful 😁

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(Original post by

Yep it's correct now.

**RDKGames**)Yep it's correct now.

This time it concerns using differentiation to find equations of the tangent and the normal.

a) Find the equation of the tangent to the curve f(x)=-x^2-5x+2 at the point where x=1:

To solve this I found the derivative of the function; dy/dx = -2x-5

Then I found the derivative when x=1, dy/dx=-2(1)=5=-7

When x=1, y= -(1)^2-5(1)+2=-4

The tangent is thus the line passing through (1,-4) with a gradient of -7.

Therefore, the equation of the tangent is y-(-4)/x-1=-7

y-(-4)=-7x-7

y=-7x-11

b) Find the equation of the normal to the curve f(x)-x^2-5x+2 at the point where x=-3

To solve this I used a similar method to above. I found the derivative of the function: dy/dx = -2x-5

At the point where x=-3, dy/dx= 1

When x=-3, y=(=3)^2=5(=3)+2=8

The normal is perpendicular to the tangent at the point (-3,8). If two lines are perpendicular then the product of their gradients is equal to -1:

m1*m2=-1

Therefore, 1 *m2=-1

m2= -1

The gradient of the normal at (-3,8) will be -1.

The equation of the normal is y-8/x-(-3)=-1

y-8=-x-3

y=5-x

c) Find the coordinates of P, the point where the tangent and normal meet.

This is where I am more confused since would this be point (-3,8)? Or are my above calculations incorrect? I really appreciate all of your help and guidance, have a wonderful evening 😁

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a) Find the equation of the tangent to the curve f(x)=-x^2-5x+2 at the point where x=1:

Therefore, the equation of the tangent is y-(-4)/x-1=-7

y-(-4)=-7x-7

**LukeWatson4590**)a) Find the equation of the tangent to the curve f(x)=-x^2-5x+2 at the point where x=1:

Therefore, the equation of the tangent is y-(-4)/x-1=-7

y-(-4)=-7x-7

b) looks fine.

For part c), I suspect they're looking for the point where the tangent from part a) intersects the normal from part b). So, you need to do a bit of simultaneous equations.

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(Original post by

For part a), you made a slip towards the end.

b) looks fine.

For part c), I suspect they're looking for the point where the tangent from part a) intersects the normal from part b). So, you need to do a bit of simultaneous equations.

**ghostwalker**)For part a), you made a slip towards the end.

b) looks fine.

For part c), I suspect they're looking for the point where the tangent from part a) intersects the normal from part b). So, you need to do a bit of simultaneous equations.

a) Sorry I had not realised. In which case; -7(x-1)=-7x+7

y=-7x+7-4

y=-7x+3?

c) I apologise but I am still struggling a little here. Do you mean where the tangent of f(x)=-x^2-5x+2 at the point where x=1 intersects with the normal of f(x)=-x^2-5x+2 at the point where x=-3. And in which case, how would I approach solving the simultaneous equations?

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(Original post by

Hello ghostwalker, thank you for your reply and your explanation.

a) Sorry I had not realised. In which case; -7(x-1)=-7x+7

y=-7x+7-4

y=-7x+3?

c) I apologise but I am still struggling a little here. Do you mean where the tangent of f(x)=-x^2-5x+2 at the point where x=1 intersects with the normal of f(x)=-x^2-5x+2 at the point where x=-3. And in which case, how would I approach solving the simultaneous equations?

**LukeWatson4590**)Hello ghostwalker, thank you for your reply and your explanation.

a) Sorry I had not realised. In which case; -7(x-1)=-7x+7

y=-7x+7-4

y=-7x+3?

c) I apologise but I am still struggling a little here. Do you mean where the tangent of f(x)=-x^2-5x+2 at the point where x=1 intersects with the normal of f(x)=-x^2-5x+2 at the point where x=-3. And in which case, how would I approach solving the simultaneous equations?

a) yes,

c) Yes. They're standard simultaneous equations:

y= -7x+3

y=5-x

In this case just equating them to eliminate y would be the way to go. Solve for x, sub back in and find the y value.

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(Original post by

a) yes,

c) Yes. They're standard simultaneous equations:

y= -7x+3

y=5-x

In this case just equating them to eliminate y would be the way to go. Solve for x, sub back in and find the y value.

**ghostwalker**)a) yes,

c) Yes. They're standard simultaneous equations:

y= -7x+3

y=5-x

In this case just equating them to eliminate y would be the way to go. Solve for x, sub back in and find the y value.

Rearrange both equations:

y+x=5

y+7x=3

Subtract the equations:

-6x=2

x=2/-6

x=-1/3

When x=-1/3; y+(-1/3)=5

y=5+1/3

y=16/3

So the coordinates of the intersection are (-1/3,16/3). Would this be correct? 😁

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(Original post by

Oh right. In order to solve:

Rearrange both equations:

y+x=5

y+7x=3

Subtract the equations:

-6x=2

x=2/-6

x=-1/3

When x=-1/3; y+(-1/3)=5

y=5+1/3

y=16/3

So the coordinates of the intersection are (-1/3,16/3). Would this be correct? 😁

**LukeWatson4590**)Oh right. In order to solve:

Rearrange both equations:

y+x=5

y+7x=3

Subtract the equations:

-6x=2

x=2/-6

x=-1/3

When x=-1/3; y+(-1/3)=5

y=5+1/3

y=16/3

So the coordinates of the intersection are (-1/3,16/3). Would this be correct? 😁

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