Determine the critical numbers

[wind]

Hi I am having trouble with this problem.

Determine the critical numbers of each function and apply the first derivative test to determine whether the critical numbers correspond to local maximum or local minimum vales. Determine the local maxi mun or local minimum values.

y= x^2 - root x
y'=2x - ½x^-½
0=2x - ½x^-½

how would you solve for x? I know how to do the rest from there

thanks.

 

[stapel]

y'=2x - ½x^-½
0=2x - ½x^-½

how would you solve for x?

You have:

. . . . .0 = 2x - (1/2)x^-1/2

. . . . .0 = 2x - 1/(2 sqrt[x])

. . . . .0 = (4x sqrt[x] - 1) / (2 sqrt[x])

A fraction is zero when its numerator is zero, so:

. . . . .0 = 4x sqrt[x] - 1

. . . . .1 = 4x sqrt[x]

. . . . .1 = 16x² x

. . . . .1 = 16x³

. . . . .1/16 = x³

. . . . .4/64 = x³

. . . . .cbrt[4] / 4 = x

Note: The conversion from "1/16" to "4/64" was done solely to simplify the result upon taking the cube root.

Eliz.

 

[wind]

Thanks, but what happened during these two lines, other than that I understand.


0 = 2x - 1/(2 sqrt[x])

0 = (4x sqrt[x] - 1) / (2 sqrt[x])

did you mulitiply the top by (2 sqrt[x])? why?

Thanks

 

[skeeter]

\(\displaystyle \L 2x - \frac{1}{2\sqrt{x}}\)

common denominator is \(\displaystyle \L 2\sqrt{x}\) ...

\(\displaystyle \L \frac{2x(2\sqrt{x})}{2\sqrt{x}} - \frac{1}{2\sqrt{x}} = \frac{4x\sqrt{x} - 1}{2\sqrt{x}}\)

 

[wind]

thanks skeeter, stapel