Example 6: Use the Graphing Strategy Use the graphing strategy and analyze the function $$ f(x)=x^{4}+4 x^{3} $$. State all the pertinent information and sketch the graph of $$ f $$.

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**1. Factor the Function** We are given $$ f(x)=x^{4}+4x^{3}. $$ Factor out the common term: $$ f(x)=x^{3}(x+4). $$ **2. Find the Zeros** Setting $$ f(x)=0 $$: $$ x^{3}(x+4)=0. $$ This yields the solutions: - $$ x^{3}=0 $$ ⟹ $$ x=0 $$ (with multiplicity 3) - $$ x+4=0 $$ ⟹ $$ x=-4 $$ Thus, the $$ x $$-intercepts are at $$ (-4, 0) $$ and $$ (0, 0) $$. **3. End Behavior** The highest power is $$ x^4 $$ with a positive coefficient. Therefore: - As $$ x\to \infty $$, $$ f(x)\to \infty $$. - As $$ x\to -\infty $$, $$ f(x)\to \infty $$. **4. First Derivative and Critical Points** Differentiate using the power rule: $$ f'(x)=\frac{d}{dx}\left(x^4+4x^3\right)=4x^3+12x^2. $$ Factor the derivative: $$ f'(x)=4x^2(x+3). $$ Set $$ f'(x)=0 $$: $$ 4x^2(x+3)=0. $$ Thus, the critical points are: - $$ x=0 $$ - $$ x=-3 $$ **5. Determine Increasing/Decreasing Intervals** Test sign of $$ f'(x)=4x^2(x+3) $$ in the intervals defined by the critical points: - For $$ x<-3 $$: Choose $$ x=-4 $$. Then, $$ f'(-4)=4(16)(-4+3)=4(16)(-1)=-64<0. $$ $$ f $$ is decreasing on $$ (-\infty,-3) $$. - For $$ -30. \] $$ f $$ is increasing on $$ (-3,0) $$. - For $$ x>0 $$: Choose $$ x=1 $$. Then, $$ f'(1)=4(1)(1+3)=16>0. $$ $$ f $$ is increasing on $$ (0,\infty) $$. Thus, there is a sign change from negative to positive at $$ x=-3 $$, indicating a local minimum at $$ x=-3 $$. At $$ x=0 $$, although $$ f'(0)=0 $$, there is no change in sign (the derivative remains positive on the right), so $$ x=0 $$ is not a local extremum—it is, in fact, a point of inflection due to the multiplicity of the root. **6. Local Extrema** Calculate $$ f $$ at $$ x=-3 $$: $$ f(-3)=(-3)^4+4(-3)^3=81-108=-27. $$ Thus, there is a local minimum at $$ (-3,-27) $$. **7. Second Derivative and Concavity** Differentiate $$ f'(x) $$ to obtain the second derivative: $$ f'(x)=4x^3+12x^2, $$ $$ f''(x)=\frac{d}{dx}(4x^3+12x^2)=12x^2+24x. $$ Factor $$ f''(x) $$: $$ f''(x)=12x(x+2). $$ Set $$ f''(x)=0 $$ to find potential inflection points: $$ 12x(x+2)=0 \quad \Longrightarrow \quad x=0 \quad \text{or} \quad x=-2. $$ Determine intervals of concavity by testing values in the intervals determined by $$ x=-2 $$ and $$ x=0 $$: - For $$ x<-2 $$: Choose $$ x=-3 $$: $$ f''(-3)=12(-3)(-3+2)=12(-3)(-1)=36>0. $$ $$ f $$ is concave up on $$ (-\infty, -2) $$. - For $$ -20 $$: Choose $$ x=1 $$: $$ f''(1)=12(1)(1+2)=12(1)(3)=36>0. $$ $$ f $$ is concave up on $$ (0, \infty) $$. Thus, inflection points occur at: - $$ x=-2 $$: Compute $$ f(-2)=(-2)^4+4(-2)^3=16-32=-16 $$. Inflection point at $$ (-2,-16) $$. - $$ x=0 $$: $$ f(0)=0 $$. Inflection point at $$ (0,0) $$. **8. Summary of Properties** - **Domain:** All real numbers, $$ (-\infty,\infty) $$. - **Zeros (x-intercepts):** $$ (-4,0) $$ and $$ (0,0) $$ (with $$ x=0 $$ being a triple root). - **y-intercept:** $$ (0,0) $$. - **End Behavior:** - As $$ x\to \pm\infty $$, $$ f(x)\to \infty $$. - **Critical Points:** - $$ x=-3 $$ with $$ f(-3)=-27 $$ (local minimum). - $$ x=0 $$, which is a stationary inflection point. - **Increasing/Decreasing Intervals:** - $$ f $$ is decreasing on $$ (-\infty, -3) $$ and increasing on $$ (-3, \infty) $$. - **Concavity and Inflection Points:** - Concave up on $$ (-\infty,-2) $$ and $$ (0,\infty) $$. - Concave down on $$ (-2,0) $$. - Inflection points at $$ (-2,-16) $$ and $$ (0,0) $$. **9. Graph Sketch Description** - For $$ x\to -\infty $$, the graph rises toward $$ +\infty $$ (concave up). - It then decreases until $$ x=-3 $$, reaching a local minimum $$ (-3,-27) $$. - After $$ x=-3 $$, the function increases. - The graph crosses the $$ x $$-axis at $$ (-4,0) $$ and $$ (0,0) $$. Note that the crossing at $$ (0,0) $$ is flatter due to the triple root and corresponds to an inflection point. - Between $$ x=-2 $$ and $$ x=0 $$, the graph is concave down. - For $$ x>0 $$, the graph is increasing and becomes concave up as it proceeds to $$ +\infty $$. All these features together give a clear picture of $$ f(x)=x^4+4x^3 $$ using the graphing strategy. **Summary of the Function $$ f(x) = x^4 + 4x^3 $$:** - **Zeros:** $$ x = -4 $$ and $$ x = 0 $$ (with $$ x = 0 $$ being a triple root). - **y-intercept:** $$ (0, 0) $$. - **End Behavior:** As $$ x $$ approaches $$ \pm\infty $$, $$ f(x) $$ approaches $$ +\infty $$. - **Critical Points:** - Local minimum at $$ (-3, -27) $$. - Inflection point at $$ (0, 0) $$. - **Intervals of Increase/Decrease:** - Decreasing on $$ (-\infty, -3) $$. - Increasing on $$ (-3, \infty) $$. - **Concavity:** - Concave up on $$ (-\infty, -2) $$ and $$ (0, \infty) $$. - Concave down on $$ (-2, 0) $$. - **Inflection Points:** - $$ (-2, -16) $$. - $$ (0, 0) $$. **Graph Description:** - The graph starts from $$ -\infty $$ and rises to a local minimum at $$ (-3, -27) $$. - It then increases, crossing the x-axis at $$ (-4, 0) $$ and $$ (0, 0) $$. - Between $$ x = -2 $$ and $$ x = 0 $$, the graph is concave down. - For $$ x > 0 $$, the graph continues to rise and becomes concave up. This comprehensive analysis provides a clear understanding of the function's behavior using the graphing strategy.

Example 6: Use the Graphing Strategy Use the graphing strategy and analyze the function \( f(x)=x^{4}+4 x^{3} \). State all the pertinent information and sketch the graph of \( f \).

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