Suppose f is differentiable on (-infinity, infinity) and assume it has a local e

Question

Suppose f is differentiable on (-infinity, infinity) and assume it has a local extreme value at the point x = 8 where f(8) = 0. Let g(x) = xf(x) + 10 and let h(x) = xf(x) + x + 10 for all values of x. a. Evaluate g(8), h(8), g'(8), and h'(8). b. Does either g or h have a local extreme value at x = 8? Explain. a. g(8) = (Simplify your answer.) h(8) = (Simplify your answer.) g'(8) = Simplify your answer.) h'(8) = (Simplify your answer.) b. Choose the correct answer below. A. h could have a local extreme value at x = 8, but g does not. B. Both g and h could have a local extreme value at x = 8. C. Neither g nor h has a local extreme value at x = 8. D. g could have a local extreme value at x = 8, but h does not.

Explanation / Answer

(a)

given x=8, f(8)=0

local extremum value at x =8 =>f '(8) =0

(a)

g(x)=xf(x) +10

g(8)=8f(8) +10

g(8)=(8*0)+10

g(8)=10

h(x)=xf(x) +x+10

h(8)=8f(8) +8+10

h(8)=(8*0)+18

h(8)=18

g(x)=xf(x) +10

g'(x)=((1*f(x)) +(x*f '(x)))+0

g'(x)=(f(x)) +(x*f '(x))

g'(8)=(f(8)) +(8*f '(8))

g'(8)=0 +(8*f '(8))

g'(8)=0 +(8*0)

g'(8)=0

h(x)=xf(x) +x+10

h'(x)=((1*f(x)) +(x*f '(x)))+1+0

h'(x)=(f(x)) +(x*f '(x))+1

h'(8)=(f(8)) +(8*f '(8))+1

h'(8)=0 +(8*f '(8))+1

h'(8)=1 +(8*f '(8))

h'(8)=1 +(8*0)

h'(8)=1

(b)

D.g could have a local extreme value at x =8 ,but h does not

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