Homework Help

f(x)=(x^2-8x+7)/(9x^2+3x-5). Evaluate f'(x) @ x=5, f'(5)=?

user profile pic

jjmgingrich | Student, Undergraduate | Salutatorian

Posted February 23, 2013 at 9:47 PM via web

dislike 0 like

f(x)=(x^2-8x+7)/(9x^2+3x-5). Evaluate f'(x) @ x=5, f'(5)=?

Tagged with calculus, math

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 24, 2013 at 12:39 PM (Answer #1)

dislike 0 like

You need to evaluate derivative of the function at x = 5, hence, you first need to find derivative of the function, using quotient rule, such that:

`f'(x) = ((x^2-8x+7)'(9x^2+3x-5) - (x^2-8x+7)(9x^2+3x-5)')/((9x^2+3x-5)^2)`

`f'(x) = ((2x - 8)(9x^2+3x-5) - (x^2-8x+7)(18x + 3))/((9x^2+3x-5)^2)`

`f'(x) = (18x^3 + 6x^2 - 10x - 72x^2 - 24x + 40 - 18x^3 - 3x^2 + 144x^2 + 24x - 126x - 21)/((9x^2+3x-5)^2)`

Reducing duplicate factors yields:

`f'(x) = (75x^2 - 136x + 19)/((9x^2+3x-5)^2)`

You need to evaluate f'(x) at x = 5, such that:

`f'(5) = (75*5^2 - 136*5 + 19)/((9*5^2+3*5-5)^2)`

`f'(5) = (1875 - 680 + 19)/((225 + 15 - 5)^2)`

`f'(5) = 1214/(235)^2 => f'(5) = 1214/55225 ~~ 0.021`

Hence, evaluating derivative of the given function at `x = 5` , yields `f'(5) = 1214/(235)^2 ~~ 0.021.`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes