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.....

Question

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.....

anonymous · Answer

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 · Answer

See the second post please. NOT FIRST

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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