Recall the inverse function theorem; if [tex]f(x)[/tex] is locally invertible around [tex]x = a[/tex], and [tex]f(a) = b[/tex] and [tex]a = f^{-1}(b)[/tex], then
[tex]f\left(f^{-1}(x)\right) = x \implies f'\left(f^{-1}(x)\right) \left(f^{-1}\right)'(x) = 1 \\\\ \implies \left(f^{-1}\right)'(x) = \dfrac1{f'\left(f^{-1}(x)\right)} \\\\ \implies \left(f^{-1}\right)'(b) = \dfrac1{f'(a)}[/tex]
Then the given derivative value tells us that there is some value of [tex]x=a[/tex] for which [tex]f(a)=10[/tex] and [tex]f'(a) = -8[/tex]. Compute the derivative of each function and check if both of these conditions are met.
- If [tex]f(x)=-5x+15[/tex], then [tex]f'(x)=-5[/tex]. (no)
- If [tex]f(x)=-2x^3-2x+14[/tex], then [tex]f'(x)=-6x^2-2[/tex]. Then [tex]f'(a) = -6a^2-2=-8 \implies a = \pm 1[/tex]. Now, [tex]f(-1)=18[/tex] doesn't work, but [tex]f(1)=10[/tex] does. (B is the correct choice)
- If [tex]f(x)=-x^5-4x+15[/tex], then [tex]f'(x)=-5x^4 - 4[/tex]. Then [tex]f'(a)=-5a^4-4=-8 \implies a = \pm\sqrt[4]{\frac45}[/tex]. But [tex]f(a)\neq10[/tex] for either of these values. (no)
- If [tex]f(x)=e^{-2x}-x+9[/tex], then [tex]f'(x)=-2e^{-2x}-1[/tex]. Then [tex]f'(a)=-2e^{-2a}-1=-8 \implies a = \frac{\ln(2)-\ln(7)}2[/tex]. But [tex]f(a)\neq10[/tex]. (no)
