Question

find the critical numbers of the function ( f(x)=12 x^{5}-45 x^{4}-20 x^{3}-4 ) and classify them using a graph.

( x=quad ) is a

( x=quad ) is a

( x=quad ) is a

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Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=12x^{5}-45x^{4}-20x^{3}-4\) is \(f^{\prime}(x)=60x^{4}-180x^{3}-60x^{2}=60x^{2}(x^{2}-3x - 1)\).

Step2: Set the derivative equal to zero

Set \(f^{\prime}(x) = 0\). We have \(60x^{2}(x^{2}-3x - 1)=0\).
From \(60x^{2}=0\), we get \(x = 0\).
For \(x^{2}-3x - 1=0\), using the quadratic formula \(x=\frac{3\pm\sqrt{9 + 4}}{2}=\frac{3\pm\sqrt{13}}{2}\). Let \(x_1=\frac{3+\sqrt{13}}{2}\approx3.30\) and \(x_2=\frac{3-\sqrt{13}}{2}\approx - 0.30\).

Step3: Classify the critical numbers (using a graph conceptually)

• For \(x = 0\): The sign of \(f^{\prime}(x)\) does not change around \(x = 0\) (since \(f^{\prime}(x)\) has a factor of \(x^{2}\)). So \(x = 0\) is not a local maximum or minimum.
• For \(x=\frac{3-\sqrt{13}}{2}\): As \(x\) increases through \(\frac{3-\sqrt{13}}{2}\), the sign of \(f^{\prime}(x)\) changes from positive to negative. So \(x=\frac{3-\sqrt{13}}{2}\) is a local maximum.
• For \(x=\frac{3+\sqrt{13}}{2}\): As \(x\) increases through \(\frac{3+\sqrt{13}}{2}\), the sign of \(f^{\prime}(x)\) changes from negative to positive. So \(x=\frac{3+\sqrt{13}}{2}\) is a local minimum.

Answer:

\(x = 0\) is a neither a local maximum nor a local minimum.
\(x=\frac{3-\sqrt{13}}{2}\) is a local maximum.
\(x=\frac{3+\sqrt{13}}{2}\) is a local minimum.