Answer :

Answer:

Average rate of change (A(x)) of y=f(x) over an interval [a, b] is given by:

[tex]A(x) = \frac{f(b)-f(a)}{b-a}[/tex]

As per the statement:

Given:

[tex]y=f(x)=\cos (2x)[/tex] and interval [tex][0, \frac{\pi}{2}][/tex]

At x = 0

[tex]f(0) = \cos (2(0)) = \cos (0) = 1[/tex]

At [tex]x = \frac{\pi}{2}[/tex]

[tex]f(\frac{\pi}{2}) = \cos (2(\frac{\pi}{2})) = \cos (\pi) =-1[/tex]

Substitute the given values in [1] we have;

[tex]A(x) = \frac{f(\frac{\pi}{2})-f(0)}{\frac{\pi}{2}-0}[/tex]

⇒[tex]A(x) = \frac{-1-1}{\frac{\pi}{2}}[/tex]

⇒[tex]A(x) = \frac{-2}{\frac{\pi}{2}}[/tex]

⇒[tex]A(x) = \frac{-4}{\pi}[/tex]

Therefore, the  average rate of change of y=cos(2x) on the interval [0, pi/2] is, [tex]\frac{-4}{\pi}[/tex]

The average rate of change is of [tex]-\frac{4}{\pi}[/tex]

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The average rate of change of a function f(x) over an interval [a,b] is given by:

[tex]A = \frac{f(b) - f(a)}{b - a}[/tex]

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  • Interval [tex][0, \frac{\pi}{2}][/tex], thus [tex]a = 0, b = \frac{\pi}{2}[/tex]
  • The function is [tex]f(x) = \cos{2x}[/tex]

[tex]f(a) = f(0) = \cos{2(0)} = \cos{0} = 1[/tex]

[tex]f(b) = f(\frac{\pi}{2}) = \cos{2\frac{\pi}{2}} = \cos{\pi} = -1[/tex]

Then

[tex]A = \frac{-1 - 1}{\frac{\pi}{2} - 0} = -\frac{2}{\frac{\pi}{2}} = -\frac{4}{\pi}[/tex]

A similar problem is given at https://brainly.com/question/20732437

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