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Differentiation

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Original post by jacksonmeg
dont you put in the x values and the positive one is minimum, negative is max?

Essentially yes.

The description of the test


We now know that:

dydx∣x=0=250\dfrac{dy}{dx} \Big|_{x=0} = 250

dydx∣x=2=−90\dfrac{dy}{dx} \Big|_{x=2} = -90

dydx∣x=5=0\dfrac{dy}{dx} \Big|_{x=5} = 0

So you can now deduce which is the maximum and which is the minimum.

Original post by jacksonmeg
I figured it didnt matter about simplifying since you just need to put in x values, and i usually go wrong anyway. lol


I'm the exact same. That's why I got into the habit of simplifying everything so I wouldn't make any silly errors!
(edited 10 years ago)
Reply 21
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jacksonmeg
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Original post by Khallil
Essentially yes.

The description of the test


We now know that:

dydx∣x=0=250\dfrac{dy}{dx} \Big|_{x=0} = 250

dydx∣x=2=−90\dfrac{dy}{dx} \Big|_{x=2} = -90

dydx∣x=5=0\dfrac{dy}{dx} \Big|_{x=5} = 0

So you can now deduce which is the maximum and which is the minimum.



I'm the exact same. That's why I got into the habit of simplifying everything so I wouldn't make any silly errors!

I was told the numbers had to be the same, e.g. +9 and -9, is this wrong?
Original post by jacksonmeg
I was told the numbers had to be the same, e.g. +9 and -9, is this wrong?


The values for the second derivative?
If that's the case, then no. They don't have to be the same at all.

Whoever told you this must have been having a bad day.
Original post by BankOfPigs
Think it's a bit easier to do u= (5-x)^2 v = -5x^2+10x. Then you don't end up using the product rule twice.

We know that dy/dx = x(5-x)^2 [10 -5x]
Change this to: (5-x)^2 [10x-5x^2] (just multiplying in the x)
Now make u= (5-x)^2 and v= 10x-5x^2
du/dx = -2(5-x) and dv/dx = 10 -10x
Using the product rule we get
d^2 y / dx^2 = (10x-5x^2)(-10+2x) + (5-x)^2(10-10x)
Substitute in your values that you found for the turning points (x=0,5,2) into the above equation.

Can you go from here?
Original post by Khallil
x

Is my answer correct above? I ask as your answer seems very different to my answer, (I help people on TSR as a form of revision for myself).
Original post by Alex-Torres
Is my answer correct above? I ask as your answer seems very different to my answer, (I help people on TSR as a form of revision for myself).


They're both the same. I just factored out 1010 and (5−x)(5-x).

d2ydx2=Your answer=(10x−5x2)(−10+2x)+(5−x)2(10−10x)=5(2x−x2)(−10+2x)+10(5−x)2(1−x)=−10(2x−x2)(5−x)+10(5−x)2(1−x)=10(5−x)(−(2x−x2)+(5−x)(1−x))=10(5−x)(x2−2x+5−6x+x2)=10(5−x)(2x2−8x+5)=My answer\begin{aligned} \dfrac{d^2y}{dx^2} & = \text{Your answer} \\ & = (10x-5x^2)(-10+2x) + (5-x)^2(10-10x) \\ & = 5(2x-x^2)(-10+2x) + 10(5-x)^2(1-x) \\ & = -10(2x-x^2)(5-x) + 10(5-x)^2(1-x) \\ & = 10(5-x) \left( -(2x-x^2) + (5-x)(1-x) \right) \\ & = 10(5-x) \left( x^2 - 2x + 5 - 6x + x^2 \right) \\ & = 10(5-x) \left( 2x^2 - 8x + 5 \right) \\ & = \text{My answer} \end{aligned}

:wink:
(edited 10 years ago)
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jacksonmeg
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Original post by Khallil
They're both the same. I just factored out 1010 and (5−x)(5-x).

d2ydx2=Your answer=(10x−5x2)(−10+2x)+(5−x)2(10−10x)=5(2x−x2)(−10+2x)+10(5−x)2(1−x)=−10(2x−x2)(5−x)+10(5−x)2(1−x)=10(5−x)(−(2x−x2)+(5−x)(1−x))=10(5−x)(x2−2x+5−6x+x2)=10(5−x)(2x2−8x+5)=My answer\begin{aligned} \dfrac{d^2y}{dx^2} & = \text{Your answer} \\ & = (10x-5x^2)(-10+2x) + (5-x)^2(10-10x) \\ & = 5(2x-x^2)(-10+2x) + 10(5-x)^2(1-x) \\ & = -10(2x-x^2)(5-x) + 10(5-x)^2(1-x) \\ & = 10(5-x) \left( -(2x-x^2) + (5-x)(1-x) \right) \\ & = 10(5-x) \left( x^2 - 2x + 5 - 6x + x^2 \right) \\ & = 10(5-x) \left( 2x^2 - 8x + 5 \right) \\ & = \text{My answer} \end{aligned}

:wink:

right, so 250 is the min, -90 the max?
Reply 27
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jacksonmeg
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Original post by Alex-Torres
Is my answer correct above? I ask as your answer seems very different to my answer, (I help people on TSR as a form of revision for myself).


Could you check this for me please?

Solve |x-2| < 2|x-1|
square both sides
(x-2)(x-2) < 2(x-1)(x-1)
x^2 -4x +4 < 2(x^2 -2x +1)
x^2 -4x +4 < 2x^2 -4x +2
0 < x^2 -2
2 < x^2
-sqrt(2) < x < sqrt(2)
Original post by jacksonmeg
Could you check this for me please?

Solve |x-2| < 2|x-1|
square both sides
(x-2)(x-2) < 2(x-1)(x-1)
x^2 -4x +4 < 2(x^2 -2x +1)
x^2 -4x +4 < 2x^2 -4x +2
0 < x^2 -2
2 < x^2
-sqrt(2) < x < sqrt(2)

Firstly, do you know what the answer is?
Reply 29
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jacksonmeg
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Original post by Alex-Torres
Firstly, do you know what the answer is?

no
Reply 30
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Liamnut
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Original post by jacksonmeg
no


Try sketching both mod graphs and looking at the intervals in which one is greater than the other if you're struggling with the squaring (IMO also harder) method.
Original post by jacksonmeg
right, so 250 is the min, -90 the max?


The point at which the second derivative is equal to 250 is the local minimum.

The point associated with a second derivative of -90 is a local maximum.

Posted from TSR Mobile
Original post by jacksonmeg
Could you check this for me please?

Solve |x-2| < 2|x-1|
square both sides
(x-2)(x-2) < 2(x-1)(x-1)
x^2 -4x +4 < 2(x^2 -2x +1)
x^2 -4x +4 < 2x^2 -4x +2
0 < x^2 -2
2 < x^2
-sqrt(2) < x < sqrt(2)

Original post by Liamnut
Try sketching both mod graphs and looking at the intervals in which one is greater than the other if you're struggling with the squaring (IMO also harder) method.


This is the best method for questions like these!
Reply 33
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jacksonmeg
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Original post by Khallil
This is the best method for questions like these!

im not sure what 2|x-1| looks like
Original post by jacksonmeg
im not sure what 2|x-1| looks like


This is just a basic transformation of the absolute value function.

Hint

(edited 10 years ago)
Reply 35
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jacksonmeg
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Original post by Khallil
If I told you that f(x)=∣x∣f(x) = |x|, f(x+a)f(x+a) is a horizontal translation to the left by aa units and af(x)af(x) is a vertical stretch of scale factor aa. What would the graph of 2∣x−1∣2|x-1| look like?

i mean, do you stretch x-1 by factor of 2 then take the modulus, or do the modulus of x-1 then stretch it? does it make a difference?
Original post by jacksonmeg
i mean, do you stretch x-1 by factor of 2 then take the modulus, or do the modulus of x-1 then stretch it? does it make a difference?


It doesn't make a difference in this case.
Reply 37
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davros
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Original post by jacksonmeg
i mean, do you stretch x-1 by factor of 2 then take the modulus, or do the modulus of x-1 then stretch it? does it make a difference?


Note that 2|x-1| = |2x - 2| which may be an easier way to visualize it :smile:
Reply 38
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jacksonmeg
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Original post by Khallil
It doesn't make a difference in this case.

is one of them x<0
Original post by jacksonmeg
is one of them x<0


:yep:

Now find the other one!

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