Quiz 7 - First derivative test and linearization

Problem 1

Question

Find the absolute maximum and minimum for the function \(y = 6\sqrt{x}-2x^{2/3}\) on the interval \(0\leq x \leq 4\).

Solution

First, we find the critical points on the interval: \[ \begin{align} y' = 0 &=3(x)^{-1/2} - \frac{4}{3}x^{-1/3} \\ \frac{4}{3}x^{-1/3} &= 3(x)^{-1/2} \\ 4 &= 9(x)^{-1/2 + 1/3} \\ \frac{4}{9} &= (x)^{-1/6} \\ (x)^{1/6} &= \frac{9}{4} \\ x &= \left(\frac{9}{4}\right)^6 \end{align} \] This is obviously outside our interval. So, we check the endpoints: \[ \begin{align} y(0) &= 0 \\ y(4) &= 12 - 4\sqrt[3]{2} \\ \end{align} \] Thus, our minimum is \(\boxed{x=0}\) and maximum is \(\boxed{x=4}\).

Problem 2

Question

Find the linearization of \(f(x) = e^{2x}\) about the point \(a=1\) the use that to approximate \(f(2)\).

Solution

The linearization form presented in the book is: \(f(x) \approx f(a) + f'(a)(x-a)\): \[ \begin{align} f(x) &\approx f(a) + f'(a)(x-a) \\ &= f(1) + f'(1)(x-1) \\ f(x) &\approx e^2 + 2e^2(x-1) \\ \Aboxed{&= 2e^2x - e^2 }\\ \end{align} \]

Using this, we have: \[ f(2) \approx e^2 + 2e^2 = \boxed{3e^2} \]