4. Let f(x) = (1/pi^3)x^3 + (9/pi)x + sinx
a) Show that f'(x) is never 0.
Attempt:
f'(x) = (3/pi^3)x^2 + (9/pi) + cosx
cos(x)'s highest y-val's are -1, 1, and (3/pi^3)x^2 + (9/pi) > 0 always
Even if,
f'(0) = 0 + (9/pi) + 1 > 0
So from examining cosine's range restrictions and the presence of a squared term, I can infer that f'(x) can never be 0. Is this a fair way of showing this?
b) Explain why f must be invertible
Not sure how I am going to go about with this one. I tried to swap x values and try to isolate for x, then interchange them to make the inverse function.....

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See the second post please. NOT FIRST

you can find the vertex of the quadratic and show that it is above 1

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