Question
Find the local max/min values of f using botht the first and second derivatie test. I can do the first but I am having trouble with the second derivative.
f(x)= x/(x^2+4)
Help please...Thank you.
f(x)= x/(x^2+4)
Help please...Thank you.
Answers
Steve
f = x/(x^2+4) = x * (x^2+4)^-1
f' = 1/(x^2+4) - x/(x^2+4)^2 * 2x
= 1/(x^2+4) - 2x^2/(x^2+4)^2
= [(x^2+4) - 2x^2]/(x^2+4)^2
= (4 - x^2)/(x^2+4)^2
f'' = -1/(x^2+4)^2 * 2x - 4x/(x^2+4)^2 + 4x^2/(x^2+4)^3 * 2x
= -6x/(x^2+4)^2 + 8x^3/(x^2+4)^3
= [-6x(x^2+4) + 8x^3]/(x^2+4)^3
= (2x^3 - 24x)/(x^2+4)^3
*whew*
Now, min/max occur when y' = 0 and y'' ≠ 0
These look pretty tough, but since the denominator of the fractions is never 0, f' and f'' are 0 when the numerator is 0.
f' = 0: 4-x^2 = 0
x = -2, 2
f'' = 0: 2x^3 - 24x = 0
x = 0, -2√3, 2√3
So, there are min/max at x = -2,2
There are inflection points at 0, -2√3, 2√3
The function grapher at
rechneronline . de / function-graphs
shows this beautifully. It even accepts the syntax used in this post. Just plug in the three functions for the three graphs. Specify x from -5 to 5 and y from -1 to 1
f' = 1/(x^2+4) - x/(x^2+4)^2 * 2x
= 1/(x^2+4) - 2x^2/(x^2+4)^2
= [(x^2+4) - 2x^2]/(x^2+4)^2
= (4 - x^2)/(x^2+4)^2
f'' = -1/(x^2+4)^2 * 2x - 4x/(x^2+4)^2 + 4x^2/(x^2+4)^3 * 2x
= -6x/(x^2+4)^2 + 8x^3/(x^2+4)^3
= [-6x(x^2+4) + 8x^3]/(x^2+4)^3
= (2x^3 - 24x)/(x^2+4)^3
*whew*
Now, min/max occur when y' = 0 and y'' ≠ 0
These look pretty tough, but since the denominator of the fractions is never 0, f' and f'' are 0 when the numerator is 0.
f' = 0: 4-x^2 = 0
x = -2, 2
f'' = 0: 2x^3 - 24x = 0
x = 0, -2√3, 2√3
So, there are min/max at x = -2,2
There are inflection points at 0, -2√3, 2√3
The function grapher at
rechneronline . de / function-graphs
shows this beautifully. It even accepts the syntax used in this post. Just plug in the three functions for the three graphs. Specify x from -5 to 5 and y from -1 to 1