Question on setting a derivative of an expression equal to 0

[minister1221]

I have a section of a problem and I am unsure which concept it is referring back to on step #3.

1. f(s) = s(1-s) / (1+s)

2. f(s) = -s² - 2s + 1 / (1+s)²

3. s² + 2s - 1 = 0

4. (1+s)² = 2

5. s = -1 + sqrt 2

What is happening on step #3? How is it that you set the derivative of f(s) = s(1-s) / (1+s) equal to 0 in that fashion?

 

[lex]

We would need to see the whole problem statement.
Your second line I think is supposed to be the derivative of [MATH]f[/MATH] wrt [MATH]s[/MATH]: i.e. [MATH]f^{'}(s)[/MATH] not [MATH]f(s)[/MATH]The third line then would be a consequence of setting [MATH]f^{'}(s)=0[/MATH], which gives an equation satisfied by the [MATH]s[/MATH] values for which the function [MATH]f(s)[/MATH] is 'stationary' (gradient=0)

 

[lookagain]

2. f(s) = -s² - 2s + 1 / (1+s)²

Your line with the derivative needs to look like this:

2. f'(s) = (-s² - 2s + 1)/(1 + s)²

 

[nasi112]

Setting a derivative to zero is used to find where the slope of the original function equal to zero. This will help you to find the maximum and minimum points on the graph.

 

[JeffM]

I have a section of a problem and I am unsure which concept it is referring back to on step #3.

1. f(s) = s(1-s) / (1+s)

2. f(s) = -s² - 2s + 1 / (1+s)²

3. s² + 2s - 1 = 0

4. (1+s)² = 2

5. s = -1 + sqrt 2

What is happening on step #3? How is it that you set the derivative of f(s) = s(1-s) / (1+s) equal to 0 in that fashion?

As has been pointed out, you seem to have failed to copy things correctly. And what you copied is missing some admittedly trivial steps.

[MATH]f(s) = \dfrac{s(1 - s)}{1 + s} = \dfrac{s - s^2}{1 + s } \implies[/MATH]
[MATH]f’(s) = \dfrac{(1 - 2s)(1 + s) - (s - s^2)(1)}{(1 + s)^2} = \dfrac{1 - s - 2s^2 - s + s^2}{(1 + s)^2} = \dfrac{1 - 2s - s^2}{(1 + s)^2}.[/MATH]
[MATH]\therefore f’(s) = 0 \implies \dfrac{1 - 2s - s^2}{(1 + s)^2} = 0 \implies[/MATH]
[MATH]- (1 + s)^2 * \dfrac{1 - 2s - s^2}{(1 + s)^2} = -(1 + s)^2 * 0 \implies s^2 + 2s - 1 = 0 \implies[/MATH]
[MATH]s = \dfrac{- 2 \pm \sqrt{2^2 - 4(1)(-1)}}{2 * 1} = \dfrac{-2 \pm \sqrt{4 + 4}}{2} =\dfrac{-2 \pm \sqrt{2 * 4}}{2} = \dfrac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}.[/MATH]