The figure below gives the behavior of the derivative of g(x) on -2 lessthanoreq

Question

The figure below gives the behavior of the derivative of g(x) on -2 lessthanorequalto x lessthanorequalto 2. Sketch a graph of g(x) and use your sketch to answer the following questions. Where does the graph of g(x) have inflection points? x = Enter your answer as a comma-separated list of values, or enter none if there are none. Where are the global maxima and minima of g on [-2, 2]? minimum at x = maximum at x = m If g(-2) = -3, what are possible values for g(0)? g(0) is in (Enter your answer as an interval, or union of intervals, giving the possible values. Thus if you know -6

Explanation / Answer

a)
g has inflection points when g'' = 0
or when g' is maximum/minimum
And this happens when x = -1

So, inf pts : x = -1

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b)
Clearly g'= 0 when x = -2 and 2

And clearly at x =-2, g' has come from positive to negative
So, at x =-2, increasing changes to decreasing
So, x = -2 is a max

And clearly when x = 2, g'is going from negative y and staying in the negative region as clearly g' is shaving the x-axis and falling back into the negative part
So, x = 2 is neither a max nor min

So, try this :
min : none
max : x = -2

If this does not work, plz try :
min : x = 2
max : x = -2

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c)
We can find that g'(x) = -(x-2)(x+2)
as this is where the g'hits the x-axis

So, g'= 4 - x^2

Integrating :
g(x) = 4x - x^3/3 + C

Using the fact that g(-2) = -3 :
-3 = 4(-2) - (-2)^3/3 + C
-3 = -8 + 8/3 + C
C = -3 + 8 - 8/3
C = 7/3

So, we have :
g(x) = 4x - x^3/3 + 7/3

Plug in x = 0 :
g(0) = 7/3
As in g(0) is in [7/3 , 7/3] -----> ANSWER

Pilug in x = 2 :
4(2) - 2^3/3 + 7/3
8 - 8/3 + 7/3
8 - 1/3
23/3

So, we have
g(2) > g(0) ----> ANSWER

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