The graph of the derivative f\' (x) of a continuous function f (x) on the integr

Question

The graph of the derivative f' (x) of a continuous function f (x) on the integral [0, 6] is shown at right below. Fill in the a f has information about the function f itself (not its derivative) f has critical numbers at x = f increases on the interval f decreases on the interval f has a local maximum at x = f has a local minimum at x = The graph of f is concave up on the interval The graph of f is concave down on the interval The x-coordinate(s) of the infection point (s) of the graph of f is (are) x =

Explanation / Answer

The graph is that of the derivative....

a)
f has criticals when f' = 0 or does not exist
the fact that the function, f is continuous means that the derivative does not exist only because there was a sharp corner at x = 3 on the function f

So, criticals are at x = 5 and 3

If this does not work, try only x = 5

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b)
f increases when f'is positive

So, increases over : [0 , 3) and (5 , 6]

And similarly decreases over (3 , 5)

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c)
At x =3, the function changes from increasing to decreasing'
So, x = 3 is a MAXIMUM

Similarly at x =5 , the function changes from decreasing to increasing
So, = 5 is a MINIMUM

Local max = 3
Local min = 5

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d)
f'' = 0 when x = 2 because this is where f' approaches a critical

Before x = 2, the function f' was decreasing
and it was thus conc down

and after x = 2, the f' is increasing
so conc up

So, conc up : (2 , 3) U (3 , 6]
conc down : [0 , 2)

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e)
Inf pts are at
x = 2

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