Asked by Kieran
Just making sure im doing things right. suppose that
f(x)=2x^2+1/ãx+1/4 x^3-x+1 Caclulate
f '(1)
So far i have 4x+(ãx^-1)+3/4 x^2-1+0
then i substituted my 1 in and came out with 4 3/4
f(x)=2x^2+1/ãx+1/4 x^3-x+1 Caclulate
f '(1)
So far i have 4x+(ãx^-1)+3/4 x^2-1+0
then i substituted my 1 in and came out with 4 3/4
Answers
Answered by
Steve
what is ã ?
also, add some parentheses. It's not clear what f(x) is
also, add some parentheses. It's not clear what f(x) is
Answered by
Kieran
sorry about that a shouldnt b in there it should be
f(x)=(2x^2)+(1/sqrt(x))+[(1/4)x^3]-x+1
the a thing was supposed to be a sqrt but i guess it doesnt copy over from work
f(x)=(2x^2)+(1/sqrt(x))+[(1/4)x^3]-x+1
the a thing was supposed to be a sqrt but i guess it doesnt copy over from work
Answered by
Reiny
Ok
f(x) = 2x^2 + 1/√x + (1/4)x^3 - x + 1
= 2x^2 + x^(-1/2) + (1/4)x^3 - x + 1
f'(x) = 4x - (1/2)x^(-3/2) + (3/4)x^2 - 1
f'(1) = 4 - (1/2)(1) + 3/4 - 1
= 13/4
f(x) = 2x^2 + 1/√x + (1/4)x^3 - x + 1
= 2x^2 + x^(-1/2) + (1/4)x^3 - x + 1
f'(x) = 4x - (1/2)x^(-3/2) + (3/4)x^2 - 1
f'(1) = 4 - (1/2)(1) + 3/4 - 1
= 13/4
Answered by
Kieran
I could be wrong but im not sure i came up with 1/4
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