Question

Let f be the function given by f(x) = x3+ 5x . For what value of x in the closed interval [1,3 ] does the instantaneous rate of change of f equal the average rate of change of f on that interval?

Answer 1

Answer:

x = 2.1

Step-by-step explanation:

Instantaneous rate of change of f = $\frac{df(x)}{dx}$

$= \frac{d}{dx}(x^{3}+5x)\\ =3x^{2} + 5$

Average rate of change of f in an interval [a,b] = $\frac{f(b)-f(a)}{b-a}$

Average rate of change = $\frac{f(3)-f(1)}{3-1} = \frac{(3^{3} + 5(3)) - (1^{3} + 5(1))}{2}$

$=\frac{(27+15) - (1 + 5)}{2}=\frac{42-6}{2}\\ \\= \frac{36}{2} = 18$

At the point when instantaneous rate of change = Average rate of change,

$3x^{2} + 5 = 18\\\\3x^{2} = 18-5 = 13$

Divide both sides by 3

$\frac{3x^{2} }{3}=\frac{13}{3}\\ \\x^{2} = 4.3333$

Take the square root of both sides

$\sqrt{x^{2} } =$ ±$\sqrt{4.3333}$

$x =$ ±2.0817

$x = 2.0817 or x = -2.0817$

But since the value of x has to be between 1 and 3,

x = 2.0817 ≅ 2.1