## Calculus with Applications (10th Edition)

Concave upward on $$(-\infty, 0 ) ,\quad (1, \infty ),$$ concave downward on $$( 0,1 )$$ inflection points at $$(0, 0), \quad (1 , -3).$$
$$f(x)=x^{\frac{8}{3}}-4 x^{\frac{5}{3}}$$ The first derivative is \begin{aligned} f^{\prime}(x) &=(\frac{8}{3})x^{\frac{5}{3}}-4(\frac{5}{3}) x^{\frac{2}{3}}\\ &=\frac{8}{3}x^{\frac{5}{3}}-\frac{20}{3} x^{\frac{2}{3}} , \end{aligned} and the second derivative is \begin{aligned} f^{\prime\prime}(x) &=\frac{8}{3} (\frac{5}{3})x^{\frac{2}{3}}-\frac{20}{3}(\frac{2}{3}) x^{\frac{-1}{3}} \\ &=\frac{40}{9} x^{\frac{2}{3}}-\frac{40}{9} x^{\frac{-1}{3}}\\ \end{aligned} We factor $f^{\prime\prime}(x)$ as \begin{aligned} f^{\prime\prime}(x) &=\frac{40}{9} x^{\frac{2}{3}}-\frac{40}{9} x^{\frac{-1}{3}}\\ &=\frac{40(x-1)}{9 x^{\frac{-1}{3}}} \end{aligned} look for an $x$-value that makes $f^{\prime\prime}(x)=0$. Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$. \begin{aligned} f^{\prime\prime}(x) =\frac{40(x-1)}{9 x^{\frac{-1}{3}}} =0 \quad \text {when } \quad x=1 \end{aligned} but also where $f^{\prime\prime}(x)$ does not exist this happens at $x=0$. So Check the sign of $f^{\prime\prime}(x)$ in the three intervals $$(-\infty, 0 ), \quad (0,1), \quad (1,\infty).$$ (1) Test a number in the interval $(-\infty, 0 )$ say $-2$ \begin{aligned} f^{\prime\prime}(-2) &=\frac{40((-2)-1)}{9 (-2)^{\frac{-1}{3}}} \\ &=\frac{40\cdot \:2^{\frac{1}{3}}}{3} \\ &\gt0 \end{aligned} to see that $f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $(-\infty, 0 )$. (2) Test a number in the interval $( 0,1)$ say $\frac{1}{2}$ \begin{aligned} f^{\prime\prime}(\frac{1}{2}) &=\frac{40((\frac{1}{2})-1)}{9 (\frac{1}{2})^{\frac{-1}{3}}} \\ &=-\frac{10\cdot \:2^{\frac{2}{3}}}{9} \\ &\lt 0 \end{aligned} to see that $f^{\prime\prime}(x)$ is negative in that interval, so $f(x)$ is concave downward on $( 0,1)$ (3) Test a number in the interval $(1, \infty )$ say $2$ \begin{aligned} f^{\prime\prime}(2) &=\frac{40((2)-1)}{9 (2)^{\frac{-1}{3}}} \\ &=\frac{40\cdot \:2^{\frac{1}{3}}}{9} \\ &\gt 0 \end{aligned} to see that $f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $(1, \infty )$. Therefore, Concave upward on $$(-\infty, 0 ) ,\quad (1, \infty ),$$ concave downward on $$( 0,1 )$$ Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $x=0$, the graph changes from concave upward to concave downward at that point, so $(0 , f(0)$ is inflection point. Also, the sign of $f^{\prime\prime}(x)$ changes from negative to positive at $x=1$, the graph changes from concave downward to concave upward at that point, so $(1 , f(1)$ is inflection point. Now ,we find $f( 0), f(1)$as follows: $$f(0)=(0)^{\frac{8}{3}}-4 (0)^{\frac{5}{3}}=0,\\ f(1)=(1)^{\frac{8}{3}}-4 (1)^{\frac{5}{3}}=-3.$$ Finally, we have inflection points at $(0, 0), (1 , -3).$