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- anonymous

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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- anonymous

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- anonymous

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, but I quickly realized that I cannot use that method to get the inverse function. I definitely need a hint for this one.
c) Show that the point (10, pi) is on the graph of the inverse fcn
If (pi, 10) is on the graph of the function f and this point exists, then can I conclude from that that the point (10,pi) exists on the graph of the inverse fcn?
d) Find the equation of the line tangent to the inverse fcn at (10, pi)
My plan was to do this:
a) Take the derivative of f
b) Find the slope of f at x = pi
c) Infer that because f(y) = x <=> f-1(x) = y, then the slope I get for this point on the function f will apply for the slope of the inverse fcn at x = 10. Is this even true?

- anonymous

See the second post please. NOT FIRST

- anonymous

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

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- anonymous

makes the function invertible because if the derivative is always positive the function is always increasing hence one to one

- anonymous

don't expect to actually find the inverse as a function

- anonymous

to show that
\[(1,\pi)\] is on the graph of the inverse, show that
\[f(\pi)=10\]

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