Answered

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Let f be the function given by f (x) = x^3 - 2x² + 5x - 16. For what value of x in the closed interval [0,5) does the instantaneous rate of change of f equal the
average rate of change of f over that interval?

Sagot :

LammettHash

f(x) is a polynomial, so it's continuous and differentiable everywhere and hence satisfies the conditions for the mean value theorem. Then there is some c in the interval (0, 5) such that

[tex]f'(c) = \dfrac{f(5) - f(0)}{5 - 0}[/tex]

We have

[tex]f(x) = x^3 - 2x^2 + 5x - 16[/tex]

and its derivative is

[tex]f'(x) = 3x^2 - 4x + 5[/tex]

Then c is such that

[tex]3c^2 - 4c + 5 = \dfrac{84-(-16)}{5-0}[/tex]

[tex]3c^2 - 4c = 15[/tex]

[tex]c^2 - \dfrac43 c = 5[/tex]

[tex]c^2 - \dfrac43 c + \dfrac49= 5 + \dfrac49[/tex]

[tex]\left(c - \dfrac23\right)^2 = \dfrac{49}9[/tex]

[tex]c - \dfrac23 = \pm\sqrt{\dfrac{49}9}[/tex]

[tex]c - \dfrac23 = \pm\dfrac73[/tex]

[tex]c = \dfrac23 \pm \dfrac73[/tex]

So f '(c) is equal to the average rate of change over [0, 5] when c = 2/3 - 7/3 = -5/3 or when c = 2/3 + 7/3 = 3. But only the second solution c = 3 falls in the given interval.

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