Graph \(f(x)=3x^{4}-8x^{3}\).
Step 1 \(f\) is a fourth-degree polynomial; its domain is all real numbers. \(f( 0) =0\), so the \(y\)-intercept is \(0\). We find the \(x\)-intercepts by solving \(f(x) =0\). \[ \begin{eqnarray*} 3x^{4}-8x^{3} &=&0 \\ x^{3}( 3x-8) &=&0 \\ x &=&0 \qquad\hbox{or}\qquad x=\dfrac{8}{3} \end{eqnarray*} \]
There are two \(x\)-intercepts: \(0\) and \(\dfrac{8}{3}\). We plot the intercepts \((0,0)\) and \(\left( \dfrac{8}{3},0\right)\)
Step 2 Polynomials have no asymptotes, but the end behavior of \(f\) resembles the power function \(y=3x^{4}\).
Step 3 \(f^\prime (x) =12x^{3}-24x^{2}=12x^{2}(x-2)\qquad f^{\prime \prime} (x) =36x^{2}-48x=12x( 3x-4) \)
For polynomials, the critical numbers occur when \(f^\prime (x) =0\). \[ \begin{eqnarray*} 12x^{2}( x-2) &=&0 \\ x &=&0 \qquad \hbox{or}\qquad x=2 \end{eqnarray*} \]
When graphing a function, piece together a sketch of the graph as you go, to ensure that the information is consistent and makes sense. If it appears to be contradictory, check your work. You may have made an error.
The critical numbers are \(0\) and \(2\). At the points \((0,0)\) and \(( 2,-16)\) , the tangent lines are horizontal. We plot these points.
Step 4 To apply the Increasing/Decreasing Function Test, we use the critical numbers \(0\) and \(2\) to form three intervals.
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| Interval | Sign of f\(^\prime (x)\) | Conclusion |
|---|---|---|
| \((-\infty ,0)\) | negative | \(f\) is decreasing on \((-\infty ,0)\) |
| (0,2) | negative | \(f\) is decreasing on \((0,2)\) |
| \((2,\infty )\) | positive | \(f\) is increasing on \((2,\infty)\) |
Step 5 We use the First Derivative Test. Based on the table above, \(f( 0) =0\) is neither a local maximum value nor a local minimum value, and \(f(2) =-16\) is a local minimum value.
Step 6 \(f^{\prime \prime} (x) =36x^{2}-48x=12x( 3x-4) ;\) the zeros of \(f^{\prime \prime}\) are \(x=0\) and \(x=\dfrac{4}{3}\). To apply the Test for Concavity, we use the numbers \(0\) and \(\dfrac{4}{3}\) to form three intervals.
| Interval | Sign of f\(^{\prime \prime}\) | Conclusion |
|---|---|---|
| \((-\infty ,0)\) | positive | \(f\) is concave up on \((-\infty ,0)\) |
| \(\left( 0,\dfrac{4}{3}\right)\) | negative | \(f\) is concave down on \(\left( 0,\dfrac{4}{3}\right)\) |
| \(\left( \dfrac{4}{3},\infty \right)\) | positive | \(f\) is concave up on \(\left( \dfrac{4}{3},\infty \right)\) |
The concavity of \(f\) changes at \(0\) and \(\dfrac{4}{3}\), so the points \( (0,0)\) and \(\left( \dfrac{4}{3},-\dfrac{256}{27}\right)\) are inflection points. We plot the inflection points.
Step 7 Using the points plotted and the information about the shape of the graph, we graph the function. See Figure 46.