Question

NEED HELP WITH ALGEBRA PLEASE!!!!!!Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval. need the answers asap please

Answer 1

F(x)=x²+3x ; [-2;3]⇒Δf/Δx=(f(3)-f(-2))/(3-(-2))=4   (A)
f(x)=3x-8 ; [4;5]⇒Δf/Δx=(f(5)-f(4))/(5-4)=3           (B)
f(x)=x²-2x ;[-3; 4]⇒Δf/Δx=(f(4)-f(-3))/(4-(-3))= -1  (C)
f(x)=x²-5 ;[-1; 1]⇒Δf/Δx=(f(1)-f(-1))/(1-(-1))= 0     (D)

answer: A, B, D, C

Answer 2

Answer with explanation:

We know that the average rate of function f(x) from x=a to x=b i.e. in the interval [a,b] is calculated as follows:

$Rate\ of\ change=\dfrac{f(b)-f(a)}{b-a}$

a)

$f(x)=x^2+3x$

interval: [-2,3]

We have:

$f(3)=3^2+3\times 3\\\\\\f(3)=9+9\\\\f(3)=18$

and

$f(-2)=(-2)^2+3\times (-2)\\\\\\f(-2)=4-6\\\\\\f(-2)=-2$

The average rate of change is calculated as:

$=\dfrac{f(3)-f(-2)}{3-(-2)}\\\\\\=\dfrac{18-(-2)}{5}\\\\\\=\dfrac{20}{5}\\\\\\=4$

Hence, Average rate of change is: 4

b)

$f(x)=3x-8$

interval: [4,5]

$f(5)=3\times 5-8\\\\\\f(5)=15-8\\\\\\f(5)=7$

and

$f(4)=3\times 4-8\\\\\\f(4)=12-8\\\\\\f(4)=4$

The average rate of change is calculated as:

$=\dfrac{f(5)-f(4)}{5-4}\\\\\\=\dfrac{7-4}{1}\\\\\\=3$

Hence, Average rate of change is: 3

c)

$f(x)=x^2-2x$

interval : [-3,4]

$f(4)=4^2-2\times 4\\\\\\f(4)=16-8\\\\\\f(4)=8$

$f(-3)=(-3)^2-2\times (-3)\\\\\\f(-3)=9+6\\\\\\f(-3)=15$

The average rate of change is calculated as:

$=\dfrac{f(4)-f(-3)}{4-(-3)}\\\\\\=\dfrac{8-15}{7}\\\\\\=\dfrac{-7}{7}=-1$

Hence, Average rate of change is: -1

d)

$f(x)=x^2-5$

interval : [-1,1]

$f(1)=1-5\\\\f(1)=-4$

and

$f(-1)=(-1)^2-5\\\\\\f(-1)=1-5\\\\\\f(-1)=-4$

The average rate of change is calculated as:

$=\dfrac{f(1)-f(-1)}{1-(-1)}\\\\\\=\dfrac{-4-(-4)}{2}\\\\\\=\dfrac{0}{2}=0$

Hence, Average rate of change is: 0

Hence, on arranging the functions from greatest to least value based on average rate of change in the specified interval is:

  $f(x)=x^2+3x$

i.e.   a < b < d < c