Ask your own question, for FREE!
Mathematics 23 Online
OpenStudy (anonymous):

PLEASE HELP !!! Calculate the average rate of change for the function f(x) = x4 + 3x3 − 5x2 + 2x − 2, from x = −1 to x = 1.

OpenStudy (anonymous):

@mathstudent55

OpenStudy (freckles):

you need to find f(1) and f(-1)

OpenStudy (freckles):

then the average rate of change is \[\frac{f(1)-f(-1)}{1-(-1)}\]

OpenStudy (anonymous):

Do I plug those in where x is?

OpenStudy (freckles):

f(1) can found by using the machine labeled f x represents the input of the machine f f(1) means we are using 1 as the input in the f machine so yes replace x with 1 to find f(1)

OpenStudy (anonymous):

Okay so it would remain the same since its 1??? @freckles

OpenStudy (freckles):

\[\text{ is } 1^4+3(1)^3-5(1)^2+2(1)-2 \text{ equal to } 1?\]

OpenStudy (freckles):

well 1^(anypower)=1 so is 1+3-5+2-2 equal to 1?

OpenStudy (anonymous):

No

OpenStudy (freckles):

\[\frac{f(1)-f(-1)}{1-(-1)} \\ =\frac{[(1)^4+3(1)^3-5(1)^2+2(1)-2]-[(-1)^4+3(-1)^3-5(-1)^2+2(-1)-2]}{1-(-1)}\]

OpenStudy (anonymous):

Woah that looks confusing!

OpenStudy (freckles):

just finding f(1) and f(-1) then pluggin it in

OpenStudy (freckles):

i just didn't simplify it for you

OpenStudy (anonymous):

Oh okay! The multiple choice is only 1 digit

OpenStudy (freckles):

you should simplify the above

OpenStudy (freckles):

what did you get for f(1)? and f(-1)?

OpenStudy (anonymous):

I jusy dint get it im really sorry! Thanks for truing to help

OpenStudy (freckles):

take f(x) and replace all the x's in it with 1's to find f(1) then replace all the x's in f with -1's to find f(-1)

OpenStudy (freckles):

you can do this

OpenStudy (freckles):

\[f(x)=x^4+3x^3-5x^2+2x-2 \\f(1)=1^4+3(1)^3-5(1)^2+2(1)-2\\ f(-1)=(-1)^4+3(-1)^3-5(-1)^2+2(-1)-2\]

OpenStudy (anonymous):

Okay hold on

OpenStudy (anonymous):

Okay is the nswer -1?

OpenStudy (anonymous):

Without using technology, describe the end behavior of f(x) = −3x38 + 7x3 − 12x + 13. Can u help with this instead? @freckles

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!