Question

for which of the following increasing functions f does (f^(-1))(20) = 1/5? a) f(x)=x + 5 b) f(x)=x^3+5x + 20 c) f(x)=x^5+5x + 14 d) f(x)=e^x+5x + 19

Explanation:

Step1: Recall the formula for the derivative of an inverse - function

The formula for the derivative of the inverse function is \((f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}\). We want \((f^{-1})'(20)=\frac{1}{5}\), so we need \(f'(f^{-1}(20)) = 5\).

Step2: Find the derivative of each function

• For \(A\): \(f(x)=x + 5\), then \(f'(x)=1\).
• For \(B\): \(f(x)=x^{3}+5x + 20\), using the power - rule \((x^{n})'=nx^{n - 1}\), \(f'(x)=3x^{2}+5\).
• For \(C\): \(f(x)=x^{5}+5x + 14\), \(f'(x)=5x^{4}+5\).
• For \(D\): \(f(x)=e^{x}+5x + 19\), \(f'(x)=e^{x}+5\).

Step3: Check the value of \(x\) such that \(f(x)=20\) and \(f'(x)=5\)

• For \(A\): If \(f(x)=x + 5=20\), then \(x = 15\) and \(f'(x)=1

eq5\).

• For \(B\): Let \(f(x)=x^{3}+5x + 20 = 20\), then \(x^{3}+5x=x(x^{2}+5)=0\), so \(x = 0\). And \(f'(0)=3\times0^{2}+5 = 5\).
• For \(C\): Let \(f(x)=x^{5}+5x + 14 = 20\), \(x^{5}+5x-6 = 0\). By inspection, when \(x = 1\), \(1^{5}+5\times1-6=0\). And \(f'(1)=5\times1^{4}+5=10

eq5\).

• For \(D\): Let \(f(x)=e^{x}+5x + 19 = 20\), \(e^{x}+5x - 1=0\). The function \(y = e^{x}+5x - 1\) is increasing (\(y'=e^{x}+5>0\)). When \(x = 0\), \(e^{0}+5\times0 - 1=0\). And \(f'(0)=e^{0}+5=6

eq5\).

Answer:

B. \(f(x)=x^{3}+5x + 20\)