Function is differentiable and strictly increasing on an open interval containing
MATHEMATICS · Single correct · AP · Calculus AB · Differentiation: Composite, Implicit, and Inverse Functions · Differentiation: Composite, Implicit, and Inverse Functions
Problem
Function g is differentiable and strictly increasing on an open interval containing x=3. Given g(3)=7 and g′(3)=0, which statement best explains why (g−1)′(7) does not exist?
(a) Because g′(3)=0, the inverse has a vertical tangent at y=7, so (g−1)′(7) exists and equals ∞. (It is therefore differentiable.)
(b) Because g′(3)=0, applying (g−1)′(7)=1/g′(3) gives (g−1)′(7)=0, so the derivative exists and equals 0.
(c) Because g′(3)=0, the non-zero derivative condition for the inverse derivative is violated, so (g−1)′(7) does not exist.
(d) Since g(3)=7, swapping x and y gives (g−1)′(7)=3.