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.

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Answer:

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Answer:

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

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

Step-by-step explanation:

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

[tex]\displaystyle f(x) = 5x-10\cos x\text{ for } -2\leq x\leq 0[/tex]

First, we should evaluate the endpoints of the interval:

[tex]\displaystyle f(-2) = 5(-2) - 10\cos (-2) \approx -5.8385[/tex]

And:

[tex]f(0) = 5(0) -10\cos (0) = -10[/tex]

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:

[tex]\displaystyle f'(x) = \frac{d}{dx}\left[ 5x - 10\cos x\right][/tex]

Differentiate:

[tex]\displaystyle f'(x) = 5 + 10\sin x[/tex]

Set the function equal to zero:

[tex]\displaystyle 0 = 5+10\sin x[/tex]

And solve for x:

[tex]\displaystyle \sin x = -\frac{1}{2}[/tex]

Using the unit circle, our solutions are:

[tex]\displaystyle x = \frac{7\pi}{6} + 2n\pi\text{ or } \frac{11\pi}{6} + 2n\pi \text{ where } n\in \mathbb{Z}[/tex]

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

[tex]\displaystyle x = \frac{11\pi}{6} - 2\pi = -\frac{\pi}{6}\approx -0.5236[/tex]

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

Substituting this back into the function yields:

[tex]\displaystyle\begin{aligned} f\left(-\frac{\pi}{6}\right) &= 5\left(-\frac{\pi}{6}\right) - 10\cos \left(-\frac{\pi}{6}\right) \\ \\ &=-\frac{5\pi}{6} - 5\sqrt{3}\\ \\ &\approx -11.2782 \end{aligned}[/tex]

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.

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