Question

Let f be the function defined by 1 2 1 1 4 3 2 f x( ) = x − x + x − x 4 3 2 2 . For how many values of x in the open interval (0, 1.565) is the instantaneous rate of change of f equal to the average rate of change of f on the closed interval [0, 1.565] ?

Answer 1

There are three values of x where the instantaneous rate of change of f(x) are equal in the open and closed intervals

How to determine the number of x values?

The function is given as:

$f(x) = \frac 14x^4 - \frac 23x^3 + \frac 12x^2 - \frac 12x$

Calculate f(0)

$f(0) = \frac 14 * 0^4 - \frac 23 * 0^3 + \frac 12 *0^2 - \frac 12 * 0$

$f(0) = 0$

Calculate f(1.565)

$f(1.565) = \frac 14 * 1.565^4 - \frac 23 * 1.565^3 + \frac 12 *1.565^2 - \frac 12 * 1.565$

$f(1.565) = -0.613$

The average rate of change is then calculated using:

$m = \frac{f(1.565) - f(0)}{1.565 - 0}$

This gives

$m = \frac{-0.613 - 0}{1.565 - 0}$

$m = -0.392$

Differentiate the function f(x)

$f'(x) = x^3 - 2x^2 + x - \frac 12$

Set f'(x) to $m = -0.392$

$x^3 - 2x^2 + x - \frac 12 = -0.392$

From the graph of the above equation (see attachment), we have three values of x at:

x = 0.1492, x = 0.5614 and x = 1.2894

Hence, there are three values of x in the open interval (0,1.565)

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