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anonymous
 4 years ago
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 yval'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
 4 years ago
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 yval'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
 4 years ago
Best ResponseYou've already chosen the best response.04. 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 yval'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 <=> f1(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
 4 years ago
Best ResponseYou've already chosen the best response.0See the second post please. NOT FIRST

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can find the vertex of the quadratic and show that it is above 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0makes the function invertible because if the derivative is always positive the function is always increasing hence one to one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0don't expect to actually find the inverse as a function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to show that \[(1,\pi)\] is on the graph of the inverse, show that \[f(\pi)=10\]
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