3. Use the Second Derivative Test to find the location of all local extrema in the interval (-23, -7) for the given below. \frac{4(x-7)^2}{x+7} f(x) = If there is more than one local maxima or local minima, write each value x seperated by a comma. If a local maxima or local minima does not occur on the function, enter Ø in the appropiate box. Answer should be exact.

          3. Use the Second Derivative Test to find the location
of all local extrema in the interval (-23, -7) for the
given below.
\frac{4(x-7)^2}{x+7}
f(x) = 
If there is more than one local maxima or local minima,
write each value x seperated by a comma. If a local
maxima or local minima does not occur on the function,
enter Ø in the appropiate box. Answer should be exact.

3. Use the Second Derivative Test to find the location
of all local extrema in the interval (-23, -7) for the
given below.
(4(x-7)^2)/(x+7)
f(x) =
If there is more than one local maxima or local minima,
write each value x seperated by a comma. If a local
maxima or local minima does not occur on the function,
enter Ø in the appropiate box. Answer should be exact.

Added by Jennifer J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sam Stansfield

08/24/2023

Step 1

Step 1: Find the first derivative of the function f(x) using the quotient rule: f'(x) = [(2(4(x-7)(x+7) - 4(x-7)^(2))]/(x+7)^(2) f'(x) = [(8(x^2 - 49) - 4(x^2 - 14x + 49))]/(x+7)^(2) f'(x) = [(8x^2 - 392 - 4x^2 + 56x - 196)]/(x+7)^(2) f'(x) = [(4x^2 + 56x - Show more…

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Use the Second Derivative Test to find the location of all local extrema in the interval (-23,-7) for the given below. f(x) = (4(x-7)^(2))/(x+7) If there is more than one local maxima or local minima, write each value x separated by a comma. If a local maxima or local minima does not occur on the function, enter (O)/() in the appropriate box. Answer should be exact.