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# Derivative question

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Hi friends,

I trust you are all doing well?....please would someone just explain this very simple thing for me?..The question is..:

Graph $$y=f'(x)=mx^2+nx+k$$ goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.

On the answer sheet they provide more than one approach, one is this:

$$y=f'(x)=mx^2+nx+k$$

$$f'(x)=(x+{1 \over3})(x-1)$$

I do understand the +1/3 and the -1, what I do not understand is how is $$mx^2+nx+k$$ replaced by $$(x+{1 \over3})(x-1)$$

This may sound like a real daft question, but I honestly don't get this..How would I know in a test this is the first step to take?

Please if someone would kindly just put my finite brain to rest...All help is appreciated, thank you all

Apr 6, 2023

#1
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Hi Juriemagic  :)

y is the graph of the first derivate, not of the original function so you can ignore all the derivative notation

Graph  $$y=f'(x)=mx^2+nx+k$$   goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.

becomes

Graph  $$y=mx^2+nx+k$$   goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.

There are at most 2 roots becasue the highest power of x is 2

P and Q are the roots because the y values are 0   so you have

$$y=t(x+\frac{1}{3})(x-1)$$      where t is some constant.

Now you use the 3rd point  R(0,1) to find  t

$$1=t(0+\frac{1}{3})(0-1)\\ 1=t(\frac{1}{3})(-1)\\ 1=t*-\frac{1}{3}\\ t=-3$$

So the function is     $$y=-3(x+\frac{1}{3})(x-1)$$

Now expand that our to find  m, n  and k

At the end substitute P, Q and R in to check it is correct.   If not go find your, or mine, error.

Apr 6, 2023
#2
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Hello Melody!!,

Thank you for the explanation...yes, I am able to take everything further, it was just that step that threw me off. The answer I was looking for really, you gave to me, and it was that the y is the graph of the first derivative, which means it really is Y=mx^2+mx+k.

Okay, I fully have it now. So the formula we used was the y=m(x-x1)(x-x2)...Thank you..stay blessed!! juriemagic  Apr 7, 2023
#3
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Hi Melody,

How would the sum be done If it was $$F'(x)=mx^2+nx+k$$ ?

juriemagic  Apr 7, 2023
#4
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Hi Juriemagic.

You do not have enought points to get an exact answer for your senario.

(-1/3;0), Q(1;0) and R(0;1)

f(x) is a cubic so we need another point, preferably another root

$$f'(x)=mx^2+nx+k\\ integrate\\ f(x)=\frac{m}{3}x^3+\frac{n}{2}x^2+kx+t\\ y=\frac{m}{3}x^3+\frac{n}{2}x^2+kx+t\\$$

Sub in the 3 points and you can get all the eunknown constants in terms of just one on them.

OR

alternatively

$$y=g(x+\frac{1}{3})(x-1)(x-a)\\ sub \;in\; (0,1) \\ 1=g(+\frac{1}{3})(-1)(-a)\\ 1=g(\frac{a}{3})\\ g=\frac{3}{a}\\ so\\ y=\frac{3}{a}(x+\frac{1}{3})(x-1)(x-a)\\$$

Note :  I have not checked this for careless error.

Apr 8, 2023
#5
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Hi Melody,

nope that's great thanks...so you would then have to integrate...yes, I agree about the additional info one would need to do this integration. Thanks Melody, your time is always appreciated!..stay blessed..

juriemagic  Apr 9, 2023
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Always a pleasure  :)

Melody  Apr 9, 2023