Derivative question

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.  

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.