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 ?
To find the value of b for which f′(2) equals 0, we need to calculate the derivative of f(x) and then substitute x=2
To find the value of b for which f′(2) equals 0, we need to calculate the derivative of f(x) and then substitute x=2.
First, let’s find the derivative of f(x) using the product rule:
f′(x) = (x^3+bx+6)g′(x) + (3x^2+b)g(x)
Now, let’s substitute x=2 and set f′(2) to zero:
f′(2) = (2^3 + 2b + 6)g′(2) + (3(2)^2 + b)g(2) = 0
Substituting g(2) = 3 and g′(2) = -1:
(2^3 + 2b + 6)(-1) + (3(2)^2 + b)(3) = 0
Simplifying:
-(8 + 2b + 6) + (12 + b)(3) = 0
-14 – 2b + 36 + 3b = 0
Simplifying further:
b + 22 = 0
b = -22
Therefore, the value of b for which f′(2) equals zero is -22.
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