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Questions: Algebra BusinessCalculus
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Find the critical numbers of \(\displaystyle f(x)=x^{3} - \frac{3 x^{2}}{2} - 60 x - 18\).
Taking the derivative gives \(\displaystyle f'(x)=x^{2} - x - 20\). The critical numbers of the zeros of the derivative. Setting it equal to zero and solving gives \(\displaystyle x^{2} - x - 20 = 0 \iff \left(x - 5\right) \left(x + 4\right)=0\). The critical numbers are: [5, -4].
\begin{question}Find the critical numbers of $f(x)=x^{3} - \frac{3 x^{2}}{2} - 60 x - 18$. \soln{9cm}{Taking the derivative gives $f'(x)=x^{2} - x - 20$. The critical numbers of the zeros of the derivative. Setting it equal to zero and solving gives $x^{2} - x - 20 = 0 \iff \left(x - 5\right) \left(x + 4\right)=0$. The critical numbers are: [5, -4]. } \end{question}
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<p> <p>Find the critical numbers of <img class="equation_image" title=" \displaystyle f(x)=x^{3} - \frac{3 x^{2}}{2} - 60 x - 18 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3Dx%5E%7B3%7D%20-%20%5Cfrac%7B3%20x%5E%7B2%7D%7D%7B2%7D%20-%2060%20x%20-%2018%20" alt="LaTeX: \displaystyle f(x)=x^{3} - \frac{3 x^{2}}{2} - 60 x - 18 " data-equation-content=" \displaystyle f(x)=x^{3} - \frac{3 x^{2}}{2} - 60 x - 18 " /> . </p> </p>
<p> <p>Taking the derivative gives <img class="equation_image" title=" \displaystyle f'(x)=x^{2} - x - 20 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%3Dx%5E%7B2%7D%20-%20x%20-%2020%20" alt="LaTeX: \displaystyle f'(x)=x^{2} - x - 20 " data-equation-content=" \displaystyle f'(x)=x^{2} - x - 20 " /> . The critical numbers of the zeros of the derivative. Setting it equal to zero and solving gives <img class="equation_image" title=" \displaystyle x^{2} - x - 20 = 0 \iff \left(x - 5\right) \left(x + 4\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20-%20x%20-%2020%20%3D%200%20%5Ciff%20%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20%2B%204%5Cright%29%3D0%20" alt="LaTeX: \displaystyle x^{2} - x - 20 = 0 \iff \left(x - 5\right) \left(x + 4\right)=0 " data-equation-content=" \displaystyle x^{2} - x - 20 = 0 \iff \left(x - 5\right) \left(x + 4\right)=0 " /> . The critical numbers are: [5, -4]. </p> </p>