Question

find the critical numbers of the function ( f(x)=-6 x^{5}-15 x^{4}+10 x^{3}+6 ) and classify them. round your answers to three decimal places.
hint: factor the derivative by finding the greatest common factor first, then use the quadratic formula.
( x= ) is a select an answer
( x= ) is a select an answer
( x= ) is a select an answer
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Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=-6x^{5}-15x^{4}+10x^{3}+6\) is \(f^{\prime}(x)=-30x^{4}-60x^{3}+30x^{2}\). Factor out the greatest common factor \( - 30x^{2}\), so \(f^{\prime}(x)=-30x^{2}(x^{2}+2x - 1)\).

Step2: Find the critical numbers

Set \(f^{\prime}(x) = 0\).

• For \(-30x^{2}=0\), we get \(x = 0\).
• For \(x^{2}+2x - 1=0\), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 1\), \(b=2\), \(c=-1\). Then \(x=\frac{-2\pm\sqrt{4 + 4}}{2}=\frac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2}\). \(x=-1+\sqrt{2}\approx0.414\) and \(x=-1-\sqrt{2}\approx - 2.414\).

Step3: Classify the critical numbers

Use the second - derivative test. The second derivative \(f^{\prime\prime}(x)=-120x^{3}-180x^{2}+60x\).

• For \(x = 0\): \(f^{\prime\prime}(0)=0\), the second - derivative test is inconclusive. But looking at the sign of \(f^{\prime}(x)\) around \(x = 0\). For \(x\lt0\) (e.g., \(x=-0.1\)), \(f^{\prime}(-0.1)=-30\times(-0.1)^{2}[(-0.1)^{2}+2\times(-0.1)-1]=-30\times0.01\times(0.01 - 0.2 - 1)\gt0\). For \(x\gt0\) (e.g., \(x = 0.1\)), \(f^{\prime}(0.1)=-30\times(0.1)^{2}[(0.1)^{2}+2\times(0.1)-1]=-30\times0.01\times(0.01+0.2 - 1)\lt0\). Since the derivative changes sign from positive to negative at \(x = 0\), \(x = 0\) is a local maximum.
• For \(x=-2.414\): \(f^{\prime\prime}(-2.414)=-120\times(-2.414)^{3}-180\times(-2.414)^{2}+60\times(-2.414)\). \(f^{\prime\prime}(-2.414)\gt0\), so \(x=-2.414\) is a local minimum.
• For \(x = 0.414\): \(f^{\prime\prime}(0.414)=-120\times(0.414)^{3}-180\times(0.414)^{2}+60\times(0.414)\). \(f^{\prime\prime}(0.414)\lt0\), so \(x = 0.414\) is a local maximum.

Answer:

\(x=-2.414\) is a local minimum.
\(x = 0\) is a local maximum.
\(x = 0.414\) is a local maximum.