Suppose f '' is continuous on (−[infinity], [infinity]). (a) If f '(−5) = 0 and f ''(−5) = −1, what can you say about f ? At x = −5, f has a local maximum. At x = −5, f has a local minimum. At x = −5, f has neither a maximum nor a minimum. More information is needed to determine if f has a maximum or minimum at x = −5. (b) If f '(1) = 0 and f ''(1) = 0, what can you say about f ? At x = 1, f has a local maximum. At x = 1, f has a local minimum. At x = 1, f has neither a maximum nor a minimum. More information is needed to determine if f has a maximum or minimum at x = 1.

Using concepts of critical points, it is found that:

a) At x = −5, f has a local maximum.

b) At x = 1, f has neither a maximum nor a minimum.

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A critical value of a function f(x) is a value of for which: .

The second derivative test is also applied:

Item a:

At x = −5, f has a local maximum.

Item b:

At x = 1, f has neither a maximum nor a minimum.

A similar problem is given at link

Step-by-step explanation:

a) The first derivative helps considering f decreases or increases. Also, when f'(x) = 0, the function gets local max/min depends on how it acts.

The second derivative helps determining the concave up/down.

At x = -5, f"(-5) = -1 <0 That means the function f have concave down. Also, it shows f increases before -5 and decreases after -5.

Hence f'(-5) = 0 shows f gets maximum at -5.

b) At the point where f" =0, the function has a reflecting point and we need more information to determine if f has a local max/min there.

y=1/4x-4

step-by-step explanation:

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: