# How do you find the maximum, minimum and inflection points and concavity for the function #f(x) = (4x)/(x^2+4)#?

#color(red)(x=2)#

#color(red)(x=-2)#

Given -

#f(x)=(4x)/(x^2+4)#

Find the first Derivative.

#f^'(x)={[(x^2+4)(4)]-[(4x)(2x)]}/(x^2+4)^2#

#f^'(x)=(4x^2+16-8x^2)/(x^2+4)^2#

#f^'(x)=(-4x^2+16)/(x^2+4)^2#

Find the Second Derivative

#f^''(x)={[(x^2+4)^2(-8x)]-[(-4x^2+16)(2)(x^2+4)(2x)]}/[(x^2+4)^2]^2#

#f^''(x)={[(x^2+4)^2(-8x)]-[4x(-4x^2+16)(x^2+4)]}/(x^2+4)^4#

#f^''(x)={(x^2+4)[(x^2+4)(-8x)]-[4x(-4x^2+16)]}/(x^2+4)^4#

#f^''(x)={[(x^2+4)(-8x)]-[4x(-4x^2+16)]}/(x^2+4)^3#

#f^''(x)= (-8x^3-32x+16x^3-64x)/(x^2+4)^3#

#f^''(x)=(8x^3-96x)/(x^2+4)^3#

To find the Maxima and Minima, set the 1st derivative equal to zero.

#f^'(x)=0 => (-4x^2+16)/(x^2+4)^2=0#

#-4x^2+16=0#

#x^2=(-16)/(-4)=4#

#x=+-sqrt4#

#color(red)(x=2)#

#color(red)(x=-2)#

At

#f^''(x)=(8x^3-96x)/(x^2+4)^3=[8(2)^3-96(2)]/[(2)^2+4]^3=(64-192)/1728=(-128)/1728<0#

At

Hence the function has a maximum at

At

#f^''(x)=(8x^3-96x)/(x^2+4)^3=[8(-2)^3-96(-2)]/[(-2)^2+4]^3=(64+192)/1728=(256)/1728<0#

At

Hence the function has a minimum at

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