f is a function that is differentiable for all reals. The value of f ′(x) is given for several values of x in the table below.
The table:
x -8,-3,0,3,8
f'(x)-4,-2,0,4,5

If f ′(x) is always increasing, which statement about f(x) must be true?
f(x) passes through the origin
f(x) is concave down for all x
f(x) has a relative minimum at x=0
f(x) has an inflection point at x=0

My answer is the last one,d

f is a differentiable function on the interval [0,1], and g(x)=f(2x) the table below gives the values of f'(x). What is the value of g'(0.2)?
x .1,.2,.3,.4,.5
f'(x) 1,2,3,-4,5
Options:
2
-8
Cannot be determined

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1. #1 not D, since f' is always increasing, f" cannot be zero anywhere. So, no inflection points. I like C, since f'(0) = 0. f has a minimum, since f" > 0 and thus f is concave up.

g(x) = f(2x), so
g'(x) = f'(2x)*2
g'(0.2) = f'(0.2)*2 = 2*2 = 4
You are correct.

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2. Come to think of it, I changed my mind.

g'(0.2) = f'(0.4)*2 = -8

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3. Thank you!

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