Find $$ f^{\prime}(x) $$. $$ f(x)=4 x^{15}-8 x^{3}+8 $$ $$ f^{\prime}(x)= $$ Compare the graphs of $$ f $$ and $$ f^{\prime} $$ and use them to explain why your answer is reasonable. $$ f^{\prime}(x)=0 $$ when $$ f $$ Select--- $$ f^{\prime} $$ is positive when $$ f $$ Select-- $$ f^{\prime} $$ is negative when $$ f $$ Select---

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We are given $$ f(x)=4x^{15}-8x^3+8. $$ **Step 1. Differentiate $$ f(x) $$:** Differentiate term‐by‐term using the power rule: $$ \frac{d}{dx}\left(4x^{15}\right)=60x^{14}, $$ $$ \frac{d}{dx}\left(-8x^3\right)=-24x^2, $$ $$ \frac{d}{dx}(8)=0. $$ Hence, $$ f^{\prime}(x)=60x^{14}-24x^2. $$ This expression can be factored: $$ f^{\prime}(x)=12x^2\left(5x^{12}-2\right). $$ --- **Step 2. Find when $$ f^{\prime}(x)=0 $$:** We set $$ 12x^2\left(5x^{12}-2\right)=0. $$ Since $$ 12x^2=0 $$ when $$ x=0, $$ or $$ 5x^{12}-2=0 \quad \Longrightarrow \quad x^{12}=\frac{2}{5}. $$ Because the exponent 12 is even, the solutions are $$ x=\sqrt[12]{\frac{2}{5}} \quad \text{and} \quad x=-\sqrt[12]{\frac{2}{5}}. $$ Thus, $$ f^{\prime}(x)=0 $$ at $$ x=0, \quad x=\sqrt[12]{\frac{2}{5}}, \quad \text{and} \quad x=-\sqrt[12]{\frac{2}{5}}. $$ Since a zero derivative corresponds to horizontal tangents, we have: $$ f^{\prime}(x)=0 \text{ when } f \text{ has horizontal tangent lines (possible local extrema)}. $$ --- **Step 3. Determine where $$ f^{\prime}(x) $$ is positive or negative:** The factored form is $$ f^{\prime}(x)=12x^2\left(5x^{12}-2\right). $$ Notice that $$ 12x^2 $$ is always nonnegative (and is zero only at $$ x=0 $$). Therefore, the sign of $$ f^{\prime}(x) $$ for $$ x\neq 0 $$ depends entirely on $$ 5x^{12}-2 $$. - $$ 5x^{12}-2>0 $$ when $$ x^{12}>\frac{2}{5}. $$ - $$ 5x^{12}-2<0 $$ when $$ x^{12}<\frac{2}{5}. $$ Thus: - $$ f^{\prime}(x)>0 $$ when $$ x^{12}>\frac{2}{5} $$. Graphically, this means that where the absolute value of $$ x $$ is sufficiently large (i.e. $$ |x| > \sqrt[12]{\frac{2}{5}} $$), the function $$ f $$ is increasing. - $$ f^{\prime}(x)<0 $$ when $$ x^{12}<\frac{2}{5} $$. This corresponds to the interval where $$ |x| < \sqrt[12]{\frac{2}{5}} $$; here the function $$ f $$ is decreasing. --- **Step 4. Graph comparison rationale:** - The graph of $$ f $$ will have horizontal tangents (flat spots) exactly where $$ f^{\prime}(x)=0 $$; namely at $$ x=0 $$ and $$ x=\pm\sqrt[12]{\frac{2}{5}} $$. At these points, you may observe possible local minima or maxima. - Where $$ f^{\prime}(x)>0 $$, the graph of $$ f $$ is increasing. This occurs when $$ |x| > \sqrt[12]{\frac{2}{5}} $$. - Where $$ f^{\prime}(x)<0 $$, the graph of $$ f $$ is decreasing. This occurs when $$ |x| < \sqrt[12]{\frac{2}{5}} $$. These observations from the derivative agree with the typical behavior of a polynomial function: flat (horizontal) where the derivative is zero, increasing on intervals where the derivative is positive, and decreasing where it is negative. --- **Final Answers:** $$ f^{\prime}(x)=60x^{14}-24x^2=12x^2\left(5x^{12}-2\right). $$ - $$ f^{\prime}(x)=0 $$ when $$ f $$ has horizontal tangent lines, - $$ f^{\prime}(x)>0 $$ when $$ f $$ is increasing, - $$ f^{\prime}(x)<0 $$ when $$ f $$ is decreasing. $$ f^{\prime}(x) = 60x^{14} - 24x^2 $$. - $$ f^{\prime}(x) = 0 $$ when $$ f $$ has horizontal tangent lines. - $$ f^{\prime}(x) $$ is positive when $$ f $$ is increasing. - $$ f^{\prime}(x) $$ is negative when $$ f $$ is decreasing.

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