find the relative extrema

[Blitze105]

For the first two i am attempting to find the relative extrema. the 3rd i need to find all values of x, when the slope of the tangent line to f(x) is 1
I just need some one to look at these and tell me if i am even close...

 

[stapel]

For those for whom the picture of the text does not display (or does not display without scrolling), the post is as follows:

f(x) = 3x[sup:1kew8n1k]5[/sup:1kew8n1k] - 5x[sup:1kew8n1k]3[/sup:1kew8n1k]

f'(x) = 15x[sup:1kew8n1k]4[/sup:1kew8n1k] - 15x[sup:1kew8n1k]2[/sup:1kew8n1k]

15x[sup:1kew8n1k]2[/sup:1kew8n1k](x[sup:1kew8n1k]2[/sup:1kew8n1k] - 1) = 0

x = 0, ±1

(0, 0) f(1) = 3(1)[sup:1kew8n1k]5[/sup:1kew8n1k] - 5(1)[sup:1kew8n1k]3[/sup:1kew8n1k]

(1, -2) -2

(1, 2) f(-1) = 3(-1) - 5(-1)

2

f(x) = 9x[sup:1kew8n1k]4[/sup:1kew8n1k] + 4x[sup:1kew8n1k]3[/sup:1kew8n1k] - 36x[sup:1kew8n1k]2[/sup:1kew8n1k] - 24x

f'(x) = 36x[sup:1kew8n1k]3[/sup:1kew8n1k] + 12x[sup:1kew8n1k]2[/sup:1kew8n1k] - 72x - 24

0 = 12x[sup:1kew8n1k]2[/sup:1kew8n1k](3x + 1) - 24(3x + 1)

[stuff scratched out]

2 = x[sup:1kew8n1k]2[/sup:1kew8n1k]

x = ± sqrt[2]

f(x) = sqrt[x] => y = x[sup:1kew8n1k]1/2[/sup:1kew8n1k] find values of x when

y' = 1/(2x[sup:1kew8n1k]1/2[/sup:1kew8n1k]) of x when

(1/4, 1/2) slope = 1

1 = 1/(2x[sup:1kew8n1k]1/2[/sup:1kew8n1k])

1 = 2x[sup:1kew8n1k]1/2[/sup:1kew8n1k]

x[sup:1kew8n1k]1/2[/sup:1kew8n1k] = 1/2 x = 1/4

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[Blitze105]

All i did from there was state that..
(1/4, 1/2) is the point when the slope would equal 1

 

[mmm4444bot]

All i did from there was state that (1/4, 1/2) is the point [where] the slope [equals] 1

Hi Blitze:

In other words, you stated this point your paper twice. Okay.

Your work and results look correct for the first and third exercises, but you made an algebra-rule mistake on the second exercise that led you to miss one of the solutions.

You divided both sides of an equation by the factor (3x + 1).

We are not allowed to divide by zero; therefore, unless we know ahead of time that the expression 3x + 1 does not equal zero, we cannot divide by it.

IF you divide by 3x + 1, THEN you implicitly state that x = -1/3 is not a solution.

The proper way to solve that equation is to leave zero on one side and factor the other side.

\(\displaystyle 0 \;=\; 12x^2 \cdot (3x + 1) - 24 \cdot (3x + 1)\)

\(\displaystyle 0 \;=\; (3x + 1) \cdot (12x^2 - 24)\)

Can you go from here?

Cheers,

~ Mark

PS: Thank you for showing your work.