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:

[tex]\displaystyle b = -22[/tex]

Step-by-step explanation:

We are given the function:

[tex]\displaystyle f(x) = (x^3+bx+6)(g(x))[/tex]

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:

[tex]\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}[/tex]

Substitute using known values and solve for b:

[tex]\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}[/tex]

In conclusion, the value of b is -22.

The function f is given b is a constant and g is a differentiable function the value b (-8.4),

Let's start by finding the derivative of the given function f(x) with respect to x:

f(x) = (x³ + bx + 6) × g(x)

Using the product rule for differentiation, the derivative of f(x) is:

f'(x) = (x³ + bx + 6) × g'(x) + g(x) × (3x² + b)

Now we are interested in finding the value of b for which f'(2) = 0. So, let's plug in x = 2 and use the information given about g(2) and g'(2):

f'(2) = (2³ + 2b + 6) × g'(2) + g(2) × (3 × 2² + b)

Since g(2) = 3 and g'(2) = -1,

f'(2) = (8 + 2b + 6) × (-1) + 3 × (3 × 4 + b)

= (2b + 14 - 8) + 3× (12 + b)

= 2b + 6 + 36 + 3b

= 5b + 42

Now, we want to find the value of b for which f'(2) = 0:

5b + 42 = 0

Subtracting 42 from both sides:

5b = -42

Dividing both sides by 5:

b = -8.4

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Complete 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

-8.4

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