Question

Use calculus to find the absolute maximum and minimum values of the function. f(x) = 5x − 10 cos(x), −2 ≤ x ≤ 0 (a) Use a graph to find the absolute maximum and minimum values of the function to two decimal places.

Answer 1

The absolute maximum is about -5.84 at x = -2.

And the absolute minimum is about -11.28 at x = -π/6.

Explanation:

We want to find the absolute maximum and minimum values of the function:

`\displaystyle f(x) = 5x-10\cos x\text{ for } -2\leq x\leq 0`

First, we should evaluate the endpoints of the interval:

`\displaystyle f(-2) = 5(-2) - 10\cos (-2) \approx -5.8385`

And:

`f(0) = 5(0) -10\cos (0) = -10`

Recall that extrema of a function occurs at its critical points. The critical points of a function are whenever its derivative is zero or undefined.

So, find the derivative of the function:

`\displaystyle f'(x) = (d)/(dx)\left[ 5x - 10\cos x\right]`

Differentiate:

`\displaystyle f'(x) = 5 + 10\sin x`

Set the function equal to zero:

`\displaystyle 0 = 5+10\sin x`

And solve for x:

`\displaystyle \sin x = -(1)/(2)`

Using the unit circle, our solutions are:

`\displaystyle x = (7\pi)/(6) + 2n\pi\text{ or } (11\pi)/(6) + 2n\pi \text{ where } n\in \mathbb{Z}`

There is only one solution in the interval [-2, 0]:

`\displaystyle x = (11\pi)/(6) - 2\pi = -(\pi)/(6)\approx -0.5236`

Thus, we only have one critical point on the interval.

Substituting this back into the function yields:

`\displaystyle\begin{aligned} f\left(-(\pi)/(6)\right) &= 5\left(-(\pi)/(6)\right) - 10\cos \left(-(\pi)/(6)\right) \\ \\ &=-(5\pi)/(6) - 5√(3)\\ \\ &\approx -11.2782 \end{aligned}`

In conclusion, the absolute maximum value of f on the interval [-2, 0] is about -5.8385 at x = -2 and the absolute minimum value of f is about -11.2782 at x = -π/6.

We can see this from the graph below as well.