[Fusionary]

How would one find the maximum gradient of a curve?

What is the exact step(s) to finding the maximum gradient of a curve at any point.

Let's say And tell me why the necessary steps need to be performed.

[DavidH20]

First step is to differentiate the curve to find the gradient. Then take a second derivative - that is, differentiate the result. Set this to zero, solve the equation, then classify the solutions as maxima, minima or points of inflection as you would for normal turning points along the curve - either by considering points either side or taking a third derivative (this could get complicated); if it is positive it is a minimum, if it is negative it is a maximum.

The function y=x³ has no maximum gradient, as the gradient approaches infinity as x does.

[Fusionary]

Thank you. Just to clarify. You take the second derivative and set it equal to zero. But then how do you whether the x value would be a maximum gradient?

[Fusionary]

Let's take the example of How would one go about finding the maximum gradient of this curve?

[Fusionary]

Right, so you get:

[Fusionary]

[Rainingshame]

(Original post by Musical-Ocean)
Differentiate the equation.
Set this equal to 0.
Work out the values.
Now these values could correspond to maximums, minimums and inflections (in the case of x^3)
Take the 2nd differential, slot in the values you got from the first differential.
If the 2nd differential is negative ----> Maximum
positive -----> Minimum
0 ------------> inflection

a) you got the question wrong. y-x^3 has no maximum gradient, it just keeps increases/decrease to infinity and negative infinity.
b) to find the maximum point on a curve you:
i) do the first derivative.
ii)set it to 0 and solve to find the stationary point(s)
iii) find the second derivative and put the value of the stationary points back in to the second darivative.
iii) if it = 0 point of inflection, if it >0 it is a minimum, if it <0 it is a maximum.
c) y=x^3 has no maximum/minimum point, only a point of inflection.
i) dy/dx = 3x^2
ii)0=3x^2
x=0^(1/2)
=0
ii)second derivative= 6x
when x=0
second derivative =0 this implies it is a point of inflection.

[Fusionary]

Guys, I'm not talking about minimum and maximum points, I'm talking about values of x where the gradient has the maximum value!

[Fusionary]

Musical ocean, when you set the first differential equal to zero, the gradient is zero. I'm trying to find the maximum value of a gradient of a curve, that is unlikely to be zero.

[Musical-Ocean]

(Original post by Rainingshame)
a) you got the question wrong. y-x^3 has no maximum gradient, it just keeps increases/decrease to infinity and negative infinity.
b) to find the maximum point on a curve you:
i) do the first derivative.
ii)set it to 0 and solve to find the stationary point(s)
iii) find the second derivative and put the value of the stationary points back in to the second darivative.
iii) if it = 0 point of inflection, if it >0 it is a minimum, if it <0 it is a maximum.
c) y=x^3 has no maximum/minimum point, only a point of inflection.
i) dy/dx = 3x^2
ii)0=3x^2
x=0^(1/2)
=0
ii)second derivative= 6x
when x=0
second derivative =0 this implies it is a point of inflection.

Aaaaaa okay, I misunderstood the question.
But what I was initially explaining, and what your explaining in part b of your text was correct.

[Rainingshame]

(Original post by Musical-Ocean)
I suggest you the read the thread again.

i recommend you read my post again, i clearly gave you four steps to find a maximum/ minimum point. If you're problem was with differentiating in general you should have asked.
a) gave an answer to the first part.
b) the third part i can't answer.

[Fusionary]

No, I'm talking about the maximum value of a gradient a curve can have. I'm not talking about maximum points!

[Musical-Ocean]

(Original post by Rainingshame)
i recommend you read my post again, i clearly gave you four steps to find a maximum/ minimum point. If you're problem was with differentiating in general you should have asked.
a) gave an answer to the first part.
b) the third part i can't answer.

Im not the one asking questions here lol.

[james.h]

You already know that, to find extrema of a curve y=f(x), differentiate and solve for f'(x)=0.

If you want to find extrema of the gradient of a function, then you'll just take y=f'(x) and proceed as before, i.e: solve f''(x)=0.

[Fusionary]

Something I wanted to hear, thank you James for clarifying. So to find the "extrema" (as you poshly call it) of a gradient, you always set the second differential equal = 0, find the value of x and plug it back into the f'(x) differential (first differential?)

[Musical-Ocean]

(Original post by Fusionary)
No, I'm talking about the maximum value of a gradient a curve can have. I'm not talking about maximum points!

Yeah, I initially misread your Q. But the thing I was explaining was correct. I don't know what Rain is on about.

Anyways, do we need to know how to do this for c3?

[Mads_300]

I didn't think it was possible to find the maximum gradient value of a curve? For C3 anyway. What exam board are you doing?

[und]

You first differentiate your function to find the gradient function. This gives you the gradient for a value of x.

Now, you're just looking for a maximum on the gradient function. You do this just the same way you'd do any other standard 'find the maximum turning point of g(x)' question.

[Fusionary]

It sounds like it, it was on OCR (non-mei) Spec

[Mads_300]

(Original post by Fusionary)
It sounds like it, it was on OCR (non-mei) Spec

Oh, I'm doing that board too.. oops! May I see the question please?