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Find the absolute maximum and minimum values of the following function on the given interval. f(x)=3x−6cos(x), [−π,π]

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Answer:

Absolute minimum: x = -π / 6

Absolute maximum: x = π

Step-by-step explanation:

The candidates for the absolute maximum and minimum are the endpoints and the critical points of the function.

First, we evaluate the function at the endpoints.

At x = -π, we have


f(-\pi)=3(-\pi)-6\cos (-\pi)
\Rightarrow\boxed{f(-\pi)\approx-3.425}

At x = π, we have


f(\pi)=3(\pi)-6\cos (\pi)
\Rightarrow\boxed{f(\pi)\approx15.425.}

Next, we find the critical points and evaluate the function at them.

The critical points = are points where the first derivative of the function are zero.

Taking the first derivative of the function gives


(df(x))/(dx)=(d)/(dx)\lbrack3x-6\cos (x)\rbrack


\Rightarrow(df(x))/(dx)=3+6\sin (x)

Now the critical points are where df(x)/dx =0; therefore, we solve


3+6\sin (x)=0

solving for x gives


\begin{gathered} \sin (x)=-(1)/(2) \\ x=\sin ^(-1)(-(1)/(2)) \end{gathered}


x=-(\pi)/(6),\; x=-(5\pi)/(6)

on the interval [−π,π].

Now, we evaluate the function at the critical points.

At x = -π/ 6, we have


f(-(\pi)/(6))=3(-(\pi)/(6))-6\cos (-(\pi)/(6))
\boxed{f(-(\pi)/(6))\approx-6.77.}

At x = -5π/6, we have


f((-5\pi)/(6))=3(-(5\pi)/(6))-6\cos (-(5\pi)/(6))
\Rightarrow\boxed{f(-(5\pi)/(6))\approx-2.66}

Hence, our candidates for absolute extrema are


\begin{gathered} f(-\pi)\approx-3.425 \\ f(\pi)\approx15.425 \\ f(-(\pi)/(6))\approx-6.77 \\ f(-(5\pi)/(6))\approx-2.66 \end{gathered}

Looking at the above we see that the absolute maximum occurs at x = π and the absolute minimum x = -π/6.

Hence,

Absolute maximum: x = π

Absolute minimum: x = -π / 6

User LeMoisela
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