Question

How do you find the average rate of change of `f(x) = 4x^3 - 8x^2 - 3` over the interval [-5, 2]?

Answer 1

`100`

Explanation:

`"the average rate of change is the slope of the secant line"`
`"joining"`

`(-5,f(-5))" and "(2,f(2))`

`"in general "(f(b)-f(a))/(b-a)to[a,b]`

`f(b)=f(2)=32-32-3=-3`

`f(a)=f(-5)=-500-200-3=-703`

`rArr(-3-(-703))/7=100`

Answer 2

`100.`

Explanation:

The Average Rate of Change of a Function `f : RR to RR,` over

the Interval `[a,b] sub RR` is given by, `{f(b)-f(a)}/(b-a).`

Hence, the reqd. Rate`={f(2)-f(-5)}/{2-(-5)}..............(star).`

`f(x)=4x^3-8x^2-3 rArr f(b)-f(a),`

`=4b^3-8b^2-3-(4a^3-8a^2-3),`

`=4(b^3-a^3)-8(b^2-a^2),`

`=4(b-a)(b^2+ba+a^2)-8(b-a)(b+a),`

`=4(b-a){b^2+ba+a^2-2(b+a)},` so that,

`(f(b)-f(a))/(b-a)=4{b^2+ba+a^2-2(b+a)}, (bnea).`

With, `a=-5, and, b=2,` we have, the desired

`"the desired rate="4{4-10+25-2(-3)}=4*25=100.`