Find the critical numbers of the function
and classify
them. Round your answers to three decimal places.
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Ask by Bryant Bob. in the United States

Mar 19,2025

Question

Find the critical numbers of the function
and classify
them. Round your answers to three decimal places.
is a Select an answer
is a Select an answer
is a Select an answer

Ask by Bryant Bob. in the United States

Mar 19,2025

UpStudy AI · Accepted Answer

We begin by finding the derivative of $$ f(x)=-12x^5+15x^4+100x^3-1. $$ Taking the derivative yields $$ f'(x)=-60x^4+60x^3+300x^2. $$ We can factor out the common factor: $$ f'(x)= -60x^2 \left(x^2 - x - 5\right). $$ The critical numbers occur when $$ f'(x)=0 $$. Thus, we need to solve $$ -60x^2 \left(x^2 - x - 5\right)=0. $$ This product is zero when either factor is zero. 1. For $$ -60x^2=0, $$ we have $$ x^2=0\quad\Longrightarrow\quad x=0. $$ 2. For $$ x^2-x-5=0, $$ we use the quadratic formula: $$ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}, $$ with $$a=1$$, $$b=-1$$, and $$c=-5$$. Then $$ x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-5)}}{2(1)}=\frac{1\pm \sqrt{1+20}}{2}=\frac{1\pm \sqrt{21}}{2}. $$ Thus, the remaining critical numbers are: $$ x=\frac{1+\sqrt{21}}{2}\quad \text{and} \quad x=\frac{1-\sqrt{21}}{2}. $$ Rounding to three decimal places: $$ \frac{1+\sqrt{21}}{2} \approx \frac{1+4.583}{2}\approx 2.792, $$ $$ \frac{1-\sqrt{21}}{2} \approx \frac{1-4.583}{2}\approx -1.792. $$ Next, we classify each critical number by testing the sign of $$ f'(x) $$. The derivative is $$ f'(x)=-60x^2 \left(x^2-x-5\right). $$ Note that the factor $$ x^2 $$ is always nonnegative and is zero only at $$ x=0 $$. Define $$ g(x)=x^2-x-5. $$ Since $$ g(x) $$ is an upward opening quadratic, its roots are $$ x\approx -1.792 $$ and $$ x\approx 2.792 $$. Therefore, $$ g(x)<0 $$ on the interval $$ \left(\frac{1-\sqrt{21}}{2},\, \frac{1+\sqrt{21}}{2}\right) $$ and $$ g(x)>0 $$ outside this interval. Now examine the sign of $$ f'(x)=-60x^2 \,g(x) $$: - For $$ x < -1.792 $$: $$ x^2>0 $$ and $$ g(x)>0 $$ so $$ f'(x) $$ is negative. - For $$ -1.792 < x < 0 $$: $$ x^2>0 $$ and $$ g(x)<0 $$ so $$ f'(x) $$ is positive. - For $$ 0 < x < 2.792 $$: $$ x^2>0 $$ and $$ g(x)<0 $$ so $$ f'(x) $$ is positive. - For $$ x > 2.792 $$: $$ x^2>0 $$ and $$ g(x)>0 $$ so $$ f'(x) $$ is negative. Thus: 1. At $$ x\approx -1.792 $$: $$ f'(x) $$ changes from negative to positive. This indicates a **local minimum**. 2. At $$ x=0 $$: $$ f'(x) $$ is zero but does not change sign (it is positive on both sides). Thus, $$ x=0 $$ is a critical number but is **not** a local extreme (it behaves like an inflection point). 3. At $$ x\approx 2.792 $$: $$ f'(x) $$ changes from positive to negative, indicating a **local maximum**. The final answers, with three decimals of accuracy, are: $$ x\approx -1.792 $$ is a local minimum. $$ x\approx 0.000 $$ is a critical number that is neither a local maximum nor a local minimum. $$ x\approx 2.792 $$ is a local maximum. The critical numbers of the function $$ f(x)=-12x^5+15x^4+100x^3-1 $$ are: - $$ x \approx -1.792 $$ is a local minimum. - $$ x = 0.000 $$ is a critical point that is neither a local maximum nor a local minimum. - $$ x \approx 2.792 $$ is a local maximum.

Find the critical numbers of the function
and classify
them. Round your answers to three decimal places.
is a Select an answer
is a Select an answer
is a Select an answer

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Find the critical numbers of the function and classify them. Round your answers to three decimal places. is a Select an answer is a Select an answer is a Select an answer

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Find the critical numbers of the function
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is a Select an answer
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is a Select an answer