Question

If f'(x)=tan^-1(x^3-x) at how many points is the tangent line to the graph of y=f(x) parallel to the line y=2x

Answer 1

The tangent lines of f(x) are parallel to y = 2x when they're slopes are the same, i.e when f'(x) = 2. We know f'(x) = tan^-1(x^3-x) so we just let that equal to 2 and solve for the values of x.\[\tan^{-1}(x^3-x)=2 \implies x^3-x= \tan(2) \implies x^3-x-\tan(2)=0 \implies x \approx -1.552\] Do you understand? @marytheshade

Answer 2

@genius12 there are no solutions.

Answer 3

@oldrin.bataku

There is solutions, what are you talking about...I just gave you a solution -.-

Answer 4

There isn't an infinite number of solutions; there is just 1. @oldrin.bataku

Answer 5

@genius12 you're right, my bad... I was considering \(\tan^{-1}\) to be the principal branch which covers \(\left(-\frac\pi2,\frac\pi2\right)\).

Answer 6

Apparently wolframalpha says that there is no solutions? 0,o
http://www.wolframalpha.com/input/?i=-arctan%28x%5E3-x%29%3D2

But since you can write arctan(x) = y as tan(y) = x, shouldn't you still be able to solve for x that way and get a valid solution? Weird....

@oldrin.bataku

Answer 7

@genius12 wolframalpha presumes the principal branch of \(\tan^{-1}\), which will only yield something in \(\left(-\frac\pi2,\frac\pi2\right)\).