What is the average rate of change of f(x), represented by the graph, over the interval [0, 2]?

Mathematics

2 answers:

max2010maxim [7]3 years ago

8 0

Answer:

a it is 2

Step-by-step explanation:

zloy xaker [14]3 years ago

4 0

Answer:

The average rate of change of f(x) is 2

A is correct

Step-by-step explanation:

Given: The graph of f(x).

To find: Average rate of change of f(x) in interval [0,2]

Formula:

$\text{Average rate }=\dfrac{f(b)-f(a)}{b-a}$

Using graph, we find f(0) and f(2)

a = 0 , b= 2

f(0) = 4

f(2) = 8

$\text{Average rate } = \dfrac{f(2)-f(0)}{2-0}$

$\text{Average rate } = \dfrac{8-4}{2-0}$

$= 2$

Hence, The average rate of change of f(x) is 2

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Please explain this problem!!!

9966 [12]

tis a little of plain differentiation.

we know the radius of the cone is decreasing at 10 mtr/mins, or namely dr/dt = -10, decreasing, meaning is negative.

we know the volume is decreasing at a rate of 1346 mtr/mins or namely dV/dt = -1346, also negative.

so, when h = 9 and V = 307, what is dh/dt in essence.

we'll be needing the "r" value at that instant, so let's get it

$V=\cfrac{1}{3}\pi r^2 h\implies 307=\cfrac{\pi }{3}r^2(9)\implies \sqrt{\cfrac{307}{3\pi }}=r$

now let's get the derivative of the volume of the cone

$V=\cfrac{1}{3}\pi r^2 h\implies \cfrac{dV}{dt}=\cfrac{\pi }{3}\stackrel{product~rule}{ \left[ \underset{chain~rule}{2r\cdot \cfrac{dr}{dt}}\cdot h+r^2\cdot \cfrac{dh}{dt} \right]} \\\\\\ -1346=\cfrac{\pi }{3}\left[2\sqrt{\cfrac{307}{3\pi }}(-10)(9)~~+ ~~ \cfrac{307}{3\pi } \cdot \cfrac{dh}{dt}\right]$

$-\cfrac{4038}{\pi }=-\cfrac{180\sqrt{307}}{\sqrt{3\pi }}+\cfrac{307}{3\pi } \cdot \cfrac{dh}{dt}\implies \left[ -\cfrac{4038}{\pi }+\cfrac{180\sqrt{307}}{\sqrt{3\pi }} \right]\cfrac{3\pi }{307}=\cfrac{dh}{dt} \\\\\\ -\cfrac{12114}{307}+\cfrac{180\sqrt{3\pi }}{\sqrt{307}}=\cfrac{dh}{dt}\implies -7.920939735970634 \approx \cfrac{dh}{dt}$