Graph \(f(x)=4x^{1/3}-x^{4/3}\).
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.
| Interval | Sign of f\(^\prime\) | Conclusion |
|---|---|---|
| \((-\infty ,0)\) | positive | \(f\) is increasing on \((-\infty ,0)\) |
| (0,1) | positive | \(f\) is increasing on \((0,1)\) |
| \((1,\infty)\) | negative | \(f\) is decreasing on \((1,\infty )\) |
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.
| Interval | Sign of f\(^{\prime \prime}\) | Conclusion |
|---|---|---|
| \(( -\infty ,-2)\) | negative | \(f\) is concave down on the interval \(( -\infty ,-2)\) |
| (–2,0) | positive | \(f\) is concave up on the interval \(( -2,0)\) |
| \(( 0,\infty )\) | negative | \(f\) is concave down on the interval \(( 0,\infty )\) |
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.