Find the slope of the graph of the function at F(x) =4/x^2 when x=5
f(5)=4/5^2=5/25
Oh, that's all? Thanks so much!
But I need to find f'(5), not just f(5)
F(x) =4/x^2 y =4/x^2 to find the inverse you need to change the x and y then solve for y x = 4/y^2 y^2*x=4 y^2=4/x sqrt(y^2)=sqrt(4)/sqrt(x) y = 2/sqrt(x) = 2sqrt(x)/x
2*sqrt(5)/5
does that look right?
errr, I don't think so, sorry : / These are my answer choices: A. -8/5 B. 8/125 C. 8/5 D. -8/125
by f' u mean derivative or inverse
f=4/x^2 = 4x^(-2) f'=-2*4*x^-3= -8/x^3 f'(5) =-8/5^3=-8/125
Ohhh, that makes so much sense. Thanks!!
For some reason, I can never remember how to rewrite exponents that are in fractions. Or whenever a root is involved.
How would you do 3x^2+5x-6/x^2 when x=-5? My answer choices end up with a fraction with 3137 as the numerator somehow
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