Derivative

[Ryan Rigdon]

need help with finding the derivative of (25x)/(x^2 +4)

I used f'(x) = (g(x)*f'(x) - f(x)*g'(x))/(g(x)^2)


My answer is (25(4-x^2))/((x^2 + 4)^2)

Correct ?

 

[galactus]

Lookin' good

 

[Ryan Rigdon]

Lookin' good

thanx galactus.


to find my stationary point i set f'(x) = 0 and then solved for x

(25(4-x^2))/((x^2 + 4)^2) = 0

100 - 25x^2 = 0

100 = 25x^2

x = -4, 4

 

[Ryan Rigdon]

with my given I =[-3,3]

and my x = -4,4

since my x values are -4, 4 and the fact they dont lie on the given interval. my critical points are -3,3

next i use f(x) to find max & min values by plugging -3,3 into f(x)

f(x) = (25x)/(x^2 + 4) f(-3) = -75/13 f(3) = 75/13


Max Value = 75/13 Min Value = -75/13

 

[Ryan Rigdon]

Lookin' good

thanx galactus.


to find my stationary point i set f'(x) = 0 and then solved for x

(25(4-x^2))/((x^2 + 4)^2) = 0

100 - 25x^2 = 0

100 = 25x^2

x = -4, 4

100 = 25x^2 does not give you x = -4, 4 but it gives you x = -2,2 my mistake.


Now my critical points are -3,-2,2,3

plug these into f(x) = (25x)/(x^2 + 4) to find max and min values

f(-3) = -75/13 f(-2) = -25/4 f(2) = 25/4 f(3) = 75/13

Max val = 25/4

Min val = -25/4

 

[galactus]

Yep, that appears to be them.

The min is at (-2,-25/4) and the max at (2,25/4).

Be sure to specify. Just sayin'. Very good.