Question

The function f is given by f(x)=(x^{3}+bx+6)(g(x)), where b is a constant and g is a differentiable function satisfying g(2)=3 and g′(2)=−1. For what value of b is f′(2)=0?

7
10
12
22

Answer 1

Answer:

$\displaystyle b = -22$

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = (x^3+bx+6)(g(x))$

Where b is a constant and g is a differentiable function.

And we want to determine the value of b such that f'(2) = 0.

Find f' using the product rule:

$\displaystyle \begin{aligned} f'(x) & = \frac{d}{dx}\left[ (x^3+bx+6)(g(x))\right] \\ \\ & = \frac{d}{dx}\left[ x^3 + bx + 6\right] g(x) + (x^3+bx+6)\frac{d}{dx}\left[ g(x)\right] \\ \\ & = (3x^2+b)(g(x)) + (x^3 + bx + 6)(g'(x))\end{aligned}$

Substitute using known values and solve for b:

$\displaystyle \begin{aligned}f'(2) = 0 & = (3(2)^2+b)(g(2)) + ((2)^3+b(2)+6)(g'(2)) \\ \\ 0 & = (12+b)(3) + (14+2b)(-1) \\ \\ 0 & = 22 +b \\ \\ b & = -22\end{aligned}$

In conclusion, the value of b is -22.