Graph $f(x)=4x^{1/3}-x^{4/3}$.

Solution

Step 1 The domain of $f$ is all real numbers. Since $f(0) = 0$, the $y$-intercept is $0$. Now $f(x) = 0$ when $4x^{1/3}-x^{4/3}=x^{1/3}\left( 4-x\right) =0$ or when $x=0$ or $x=4.$ So, the $x$-intercepts are $0$ and $4$. Plot the intercepts $(0,0)$ and $\left( 4,0\right).$

Step 2 Since $\lim\limits_{x\rightarrow \infty }f(x) =\lim\limits_{x\rightarrow \infty }[ x^{1/3}(4-x)] =-\infty$, there is no horizontal asymptote. Since the domain of $f$ is all real numbers, there is no vertical asymptote.

Step 3 $$\begin{eqnarray*} f^\prime (x) &=&\dfrac{d}{\textit{dx}}( 4x^{1/3}-x^{4/3}) =\dfrac{4}{3}x^{-2/3}-\dfrac{4}{3}x^{1/3} = \dfrac{4}{3}\!\left( \dfrac{1}{x^{2/3}}-x^{1/3}\right)\\[5pt] &=&\dfrac{4}{3}\!\left( \dfrac{1-x}{ x^{2/3}}\right) \\[5pt] f^{\prime \prime} (x) &=&\dfrac{d}{\textit{dx}}\!\left( \dfrac{4}{ 3}x^{-2/3}-\dfrac{4}{3}x^{1/3}\right) =-\dfrac{8}{9}x^{-5/3}-\dfrac{4}{9} x^{-2/3}=-\dfrac{4}{9}\!\left( \dfrac{2}{x^{5/3}}+\dfrac{1}{x^{2/3}}\right)\\[5pt] &=&-\dfrac{4}{9}\cdot \dfrac{2+x}{x^{5/3}} \end{eqnarray*}$$

Since $f^\prime (x) =\dfrac{4}{3}\!\left( \dfrac{1-x}{ x^{2/3}}\right) =0$ when $x=1$ and $f^\prime (x)$ does not exist at $x = 0,$ the critical numbers are $0$ and $1$. At the point $(1,3)$, the tangent line to the graph is horizontal; at the point $(0,0)$, the tangent line is vertical. Plot these points.

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Step 4 To apply the Increasing/Decreasing Function Test, we use the critical numbers $0$ and $1$ to form three intervals.

Step 5 By the First Derivative Test, $f( 1) =3$ is a local maximum value and $f( 0) =0$ is not a local extreme value.

Step 6 We now test for concavity by using the numbers $-2$ and $0$ to form three intervals.

The concavity changes at $-2$ and at $0.$ Since $$f( -2) =4( -2) ^{1/3}- ( -2) ^{4/3}= 4\sqrt[3]{-2}-\sqrt[3]{16}\approx -7.56,$$

the inflection points are $( -2,-7.56)$ and $(0,0)$ . Plot the inflection points.

Step 7 The graph of $f$ is given in Figure 50.