The derivative, f'(x) = 6x^2+1, is never negative, so f(x) is monotonic, hence invertible.
f'(-2) = 6(-2)²+1 = 25
If point (-2, -26) is on the graph of function f(x), and the slope is 25 there, then (-26, -2) is on the graph of f⁻¹(x), and the slope is 1/25 there. The equation of the tangent line throught that point can be written in point-slope form as
... y +2 = (1/25)(x +26)
Answer:
43921
Step-by-step explanation:
Answer:
b
Step-by-step explanation:
For clarity, f(x)=x^2+x-6/x^3-1 should be written as:
x^2 + x - 6
f(x) = ------------------
x^3 - 1
Note that the denominator, x^3 - 1, factors as follows: (x - 1)(x^2 + x + 1). If we set these results = to 0 individually, we find that there is only one real root; it is x = 1. We cannot divide by zero, so conclude that the given function is undefined at x = 1. Thus, the function has the vertical asymptote x = 1.
Note that x^2 in the numerator is of a higher power than is x^3 in the denominator. As x grows large without bound, either + or - , this ratio x^2/x^3 approaches zero. Thus, the horizontal asymptote is y = 0.
The correct answer choice is (b).
