suppose we know that h\'(x) =f(g(x)), g(0)=1, g\'(0)=-2, and f\'(1)=3. further w

Question

suppose we know that h'(x) =f(g(x)), g(0)=1, g'(0)=-2, and f'(1)=3. further we also know that c=0 is a critical value of h(x). using the second derivative test, is h(0) a local max, local min, or neither? suppose we know that h'(x) =f(g(x)), g(0)=1, g'(0)=-2, and f'(1)=3. further we also know that c=0 is a critical value of h(x). using the second derivative test, is h(0) a local max, local min, or neither? suppose we know that h'(x) =f(g(x)), g(0)=1, g'(0)=-2, and f'(1)=3. further we also know that c=0 is a critical value of h(x). using the second derivative test, is h(0) a local max, local min, or neither? suppose we know that h'(x) =f(g(x)), g(0)=1, g'(0)=-2, and f'(1)=3. further we also know that c=0 is a critical value of h(x). using the second derivative test, is h(0) a local max, local min, or neither?

Explanation / Answer

h'(x) =f(g(x))

Deriving again by chain rule :

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

Plug in x = 0 :

h''(0) = f'(g(0)) * g'(0)

h''(0) = f'(1) * (-2)

h''(0) = 3 * -2

h''(0) = -6

So, since the second derivative is NEGATIVE, it means x = 0 is a LOCAL MAXIMUM

Local max --> ANSWER

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