Thread: Veriyfing that point given is either maximum or minimum

1. Veriyfing that point given is either maximum or minimum

Use an algebraic strategy to verify that the point given for the function is either a maximum or minimum.

f(x) = x3 - 3x; (-1,2)

The answer should be 0 yet i get another answer. This is what i did:[COLOR=rgb(0, 0, 0)]

[/COLOR] $(a+h)-(a)/h=(-1+h)-(-1)/h=(-1+h)^3-(-3(-1))/h$

$(-1+h)(-1+h)(-1+h)=(1-h-h+h^2)(-1+h)=-1+3h-3h^2+h^3$

$(-1+3h-3h^2+h^3-3/h$

"h's" cancel

$-1+3-3h+h^2-3=h^2-3h-1$

$h=0.01,h=-0.01$

$(0.01)^2-3(0.01)-1=-1.0299$

$(-0.01)^2-3(-0.01)-1=-0.9699$

$-1.0299+(-0.9699)/2=-0.9999$

It should = 0 however and the slopes should be negative right side and positive left side I believe proving it is a maximum.

2. Check the first step.
It should be

$\frac{( a + h) - a}{h}$ It becomes

$\frac{[(-1 + h)^3 - 3(-1 +h)] - [ (-1)^3 -3(-1)]}{h}$

Now proceed.

3. Yeh that's what I did. I'm new with latex so I really ment h dividing that whole part not just "a" which gives me the answer that I got

4. Simplify the numerator and see what you get.

5. Alright so I tried again hopefully correctly but in the end I still didn't get that 0 i'm looking for

[tex](-1+h)^3-3(-1+h)-[(-1)^3-3(-1)]/h[\MATH]

[tex]-1+3h-3h+3h^3-[-1+3][\MATH]

[tex]-1+1-3+3h-3h+3h^3/h[\MATH]

"h's cancel plus other numbers"

[tex]3h-3[\MATH]

[tex]3(0.01)-3 = -2.97[\MATH]

[tex]3(-0.01)-3 = -3.03[\MATH]

[tex]-2.97+(-3.03)/2 = -3[\MATH]

This time i get -3 instead of 0 and both "h's" left and right sides are negative instead of being left positive and right negative proving it to be a maximum

6. Alright after many many tries i managed to get a tangent of 0.0001. Not 0 but really really close so I guess that's good enough and I also got the right side to be negative and left side positive proving its a maximum. The problem was that I made a simple but at the same time major mistake simplifying.