Question

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

Answer 1

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