Given f(x)=2x3+x−8 verify that f(x) is invertible and, if so, find the equation of the tangent line to f−1(x) at the point where
x=−26. note that f(−2)=−26
1 answer:
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)
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