Find a and b from the function

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The discussion focuses on finding the coefficients a and b in the polynomial function f(x) = x^3 + ax^2 + bx, given that the stationary points occur at x = 2 and x = -4/3. The first derivative, f'(x) = 3x^2 + 2ax + b, leads to two equations: 12 + 4a + b = 0 and 16/3 - 8/3a + b = 0. Solving this system of equations yields the values of a and b necessary for the function's stationary points.

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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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