Relative Extremes Problem

[carmidy]

I'm supposed to find the relative extreme points for y=1/3x^3-4x^2+15x-3.
I know that I need to take the derivative of that which is x^2-8x+15 equal to 0. But for some reason it keeps saying my answer is wrong.

x^2-8x+15=0
(x-5)(x-3)=0
x=5,3

plug into the original equation:
x=5
y=1/3(5)^3-4(5)^2+15(5)-3 = 281/3
first point = (5, 281/3)

x=3
y=1/3(3)^3-4(3)^2+15(3)-3 = 15
second point = (3,15)

what am i doing wrong?

 

[arthur ohlsten]

your equation is ambiguous. If the equation is
y=1/[3x^3] -4x^2+15x-3 the your derivative is in error
y=[1/3]x^-3 -4x^2+15x-3
dy/dx= -x^-4-8x+15

if the equation is
y=1/[3x^2-4x^2+15x-3] than the above derivation is also in error, and I would rewrite it as
y=[3x^2-4x^2+15x-3]^-1 and then take the derivative

I hope this helps
Arthur

 

[BigGlenntheHeavy]

Right on Arthur.

\(\displaystyle Is \ your \ equation \ y \ = \ \frac{1}{3x^{3}}-4x^{2}+15x-3 \ or \ y \ = \ \frac{1}{3x^{3}-4x^{2}+15x-3}?\)

\(\displaystyle We \ don't \ expect \ one \ to \ be \ well \ versed \ in \ LaTex, \ but \ at \ this \ stage \ of \ your \ sophistication\)

\(\displaystyle in \ "The \ Calculus", \ you \ should \ have \ a \ basic \ grasp \ of \ grouping \ symbols.\)