MHB Find a and b from the function

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To find the values of a and b in the function f(x)=x^3+ax^2+bx with stationary points at x=2 and x=-4/3, the first derivative f'(x)=3x^2+2ax+b must equal zero at these points. This leads to two equations: 12 + 4a + b = 0 and 16/3 - 8/3a + b = 0. Solving this system of equations will yield the values of a and b. The discussion emphasizes the need for differentiation to approach the problem correctly. The thread has been moved to the Calculus subforum for better assistance.
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If f(x)=x^3+ax^2+bx has stationary points at: x= 2 and x= -4/3, find the value of a and b.

In my homework, can't work out, Thanks in advance
 
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rsyed5 said:
If f(x)=x^3+ax^2+bx has stationary points at: x= 2 and x= -4/3, find the value of a and b.

In my homework, can't work out, Thanks in advance

In the stationary points is...

$\displaystyle f^{\ '} (x) = 3\ x^{2} + 2\ a\ x + b = 0\ (1)$

... and that leads to the linear pair of equations...

$\displaystyle 12 + 4\ a + b = 0$

$\displaystyle \frac{16}{3} - \frac{8}{3}\ a + b = 0\ (2)$

Now You have to solve (2) finding a and b...

Kind regards$\chi$ $\sigma$
 
I have moved this thread to our Calculus subforum as this problem requires the use of differentiation. I'm going to hold off from helping at the moment because I see others already reading this thread. :D
 
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