## Calculus with Applications (10th Edition)

$$f(x)=x^{\frac{7}{3}}+x^{\frac{4}{3}}$$ critical numbers are: $\frac{-4}{7}$ and $0$. By Part 2 of the second derivative test, $\frac{-4}{7}$ leads to a relative maximum. By using the first derivative test, $0$ leads to a relative minimum.
$$f(x)=x^{\frac{7}{3}}+x^{\frac{4}{3}}$$ First, find the points where the derivative is $0$ Here \begin{aligned} f^{\prime}(x) &=(\frac{7}{3})x^{\frac{4}{3}}+ (\frac{4}{3})x^{\frac{1}{3}}\\ &=\frac{x^{\frac{1}{3}}}{3}(7x+4) \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x)& =\frac{x^{\frac{1}{3}}}{3}(7x+4)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ 7x+4 =0 \quad &\text {or} \quad x =0 \\ \\ \Rightarrow\quad\quad\quad\quad\quad\\ 7x =-4 \quad &\text {or} \quad x =0 \\ \Rightarrow\quad\quad\quad\quad\quad\\ x =\frac{-4}{7} \quad &\text {or} \quad x =0 \\ \end{aligned} critical numbers are: $\frac{-4}{7}$ and $0$. Now use the second derivative test. The second derivative is \begin{aligned} f^{\prime\prime}(x)&=\frac{7}{3} (\frac{4}{3})x^{\frac{1}{3}}+ \frac{4}{3} (\frac{1}{3})x^{\frac{-2}{3}} \\ &=\frac{28}{9} x^{\frac{1}{3}}+ \frac{4}{9}x^{\frac{-2}{3}}. \end{aligned} Evaluate $f^{\prime\prime}(x)$ first at $x=\frac{-4}{7}$, getting \begin{aligned} f^{\prime\prime}(\frac{-4}{7})&=\frac{28}{9} (\frac{-4}{7})^{\frac{1}{3}}+ \frac{4}{9}(\frac{-4}{7})^{\frac{-2}{3}}\\ &=-\frac{4^{\frac{1}{3}}\cdot \:7^{\frac{2}{3}}}{3}\\ & \approx -1.93626 \lt 0 \end{aligned} so that by Part 2 of the second derivative test, $\frac{-4}{7}$ leads to a relative maximum. Now, evaluate $f^{\prime\prime}(x)$ at $x=0$, getting \begin{aligned} f^{\prime\prime}(0)&=\frac{28}{9} (0)^{\frac{1}{3}}+ \frac{4}{9}(0)^{\frac{-2}{3}}\\ &=0 \end{aligned} then the test gives no information about extrema, so we can use the first derivative test. Check the sign of $f^{\prime}(x)$ in the intervals $$(\frac{-4}{7}, 0 ), \quad (0,\infty).$$ (1) Test a number in the interval $(\frac{-4}{7}, 0 )$ say $\frac{-1}{2}$: \begin{aligned} f^{\prime}(\frac{-1}{2}) &=(\frac{7}{3})(\frac{-1}{2})^{\frac{4}{3}}+ (\frac{4}{3})(\frac{-1}{2})^{\frac{1}{3}}\\ &=-\frac{1}{6\cdot \:2^{\frac{1}{3}}} \\ &\approx -0.13228 \end{aligned} to see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(\frac{-4}{7}, 0 )$ (2) Test a number in the interval $(0, \infty )$ say $1$: \begin{aligned} f^{\prime}(1) &=(\frac{7}{3})(1)^{\frac{4}{3}}+ (\frac{4}{3})(1)^{\frac{1}{3}}\\ &=\frac{11}{3} \\ &\approx 3.6666 \end{aligned} to see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(0, \infty ).$ So that by the first derivative test, $0$ leads to a relative minimum.