Answer:
x = 2.1
Step-by-step explanation:
Instantaneous rate of change of f = [tex]\frac{df(x)}{dx}[/tex]
[tex]= \frac{d}{dx}(x^{3}+5x)\\ =3x^{2} + 5[/tex]
Average rate of change of f in an interval [a,b] = [tex]\frac{f(b)-f(a)}{b-a}[/tex]
Average rate of change = [tex]\frac{f(3)-f(1)}{3-1} = \frac{(3^{3} + 5(3)) - (1^{3} + 5(1))}{2}[/tex]
[tex]=\frac{(27+15) - (1 + 5)}{2}=\frac{42-6}{2}\\ \\= \frac{36}{2} = 18[/tex]
At the point when instantaneous rate of change = Average rate of change,
[tex]3x^{2} + 5 = 18\\\\3x^{2} = 18-5 = 13[/tex]
Divide both sides by 3
[tex]\frac{3x^{2} }{3}=\frac{13}{3}\\ \\x^{2} = 4.3333[/tex]
Take the square root of both sides
[tex]\sqrt{x^{2} } =[/tex] ±[tex]\sqrt{4.3333}[/tex]
[tex]x =[/tex] ±2.0817
[tex]x = 2.0817 or x = -2.0817[/tex]
But since the value of x has to be between 1 and 3,
x = 2.0817 ≅ 2.1
[tex]x=2.0817[/tex]
Step-by-step explanation:
Given :
[tex]f(x )= x^3+5 \;\;\;\;\;\rm where ,\;x \epsilon [1,3][/tex]
Calculation :
Instantaneous rate of change of f
[tex]=\dfrac {d(f(x))}{dx}[/tex]
[tex]=3x^2 + 5[/tex]
The average rate of change of f in an interval [a,b] is given by
[tex]= \dfrac{f(b)-f(a)}{b-a}[/tex]
[tex]=\dfrac{(3^3+5\times 3)-(1^3+5\times 1)}{3-1}= 18[/tex]
A point when, instantaneous rate of change = average rate of change
[tex]3x^2+5=18[/tex]
[tex]x^2 = \dfrac {13}{3}[/tex]
[tex]x = \pm 2.0817[/tex]
But , [tex]x \;\epsilon \;[1,3][/tex]
Therefore, [tex]x = 2.0817[/tex]
For more information, refer the given below
https://brainly.com/question/24898810?referrer=searchResults
