91Ó°ÊÓ

Let \(f\) be a function defined on an interval \([a, b] .\) What conditions could you place on \(f\) to guarantee that $$\min f^{\prime} \leq \frac{f(b)-f(a)}{b-a} \leq \max f^{\prime}$$ where \(\min f^{\prime}\) and \(\max f^{\prime}\) refer to the minimum and maximum values of \(f^{\prime}\) on \([a, b]\) ? Give reasons for your answers.

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2 / 3}\left(x^{2}-4\right)$$

Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow \infty} \frac{2^{x}-3^{x}}{3^{x}+4^{x}}$$

Verify the formulas in Exercises by differentiation. $$\int(7 x-2)^{3} d x=\frac{(7 x-2)^{4}}{28}+C$$

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x \sqrt{4-x^{2}}$$

Find the open intervals on which the function \(f(x)=a x^{2}+\) \(b x+c, a \neq 0,\) is increasing and decreasing. Describe the reasoning behind your answer.

Verify the formulas in Exercises by differentiation. $$\int(3 x+5)^{-2} d x=-\frac{(3 x+5)^{-1}}{3}+C$$

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2} \sqrt{3-x}$$

Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow-\infty} \frac{2^{x}+4^{x}}{5^{x}-2^{x}}$$

Let \(f\) be differentiable at every value of \(x\) and suppose that \(f(1)=1,\) that \(f^{\prime}<0\) on \((-\infty, 1),\) and that \(f^{\prime}>0\) on \((1, \infty)\). a. Show that \(f(x) \geq 1\) for all \(x\). b. Must \(f^{\prime}(1)=0 ?\) Explain.