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

Question

Ezekiel Alexander · Accepted Answer

#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)#

#x# has two values

At #x=2#

#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 #x=2, f^'(x)=0; f^''<0#

Hence the function has a maximum at #x=2#

At #x=-2#

#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 #x=2, f^'(x)=0; f^''>0#

Hence the function has a minimum at #x=-2#

Samantha Adkison · Answer

undefined To find the maximum and minimum points of $ f(x) = \frac{4x}{x^2 + 4} $, we need to:

1. Take the first derivative of the function.
2. Set the derivative equal to zero and solve for $ x $.
3. Use the second derivative test to determine whether these critical points correspond to a maximum, minimum, or neither.

To find inflection points and concavity:

1. Take the second derivative of the function.
2. Set the second derivative equal to zero and solve for $ x $ to find potential inflection points.
3. Analyze the sign of the second derivative around these points to determine concavity.

Let's proceed with the calculations.

Given $ f(x) = \frac{4x}{x^2 + 4} $,

1. **First Derivative (f'(x)):**

$$ f'(x) = \frac{d}{dx} \left( \frac{4x}{x^2 + 4} \right) $$
$$ f'(x) = \frac{(4)(x^2 + 4) - 4x(2x)}{(x^2 + 4)^2} $$
$$ f'(x) = \frac{4(x^2 + 4 - 2x^2)}{(x^2 + 4)^2} $$
$$ f'(x) = \frac{4(4 - x^2)}{(x^2 + 4)^2} $$

2. **Critical Points:**

Set $ f'(x) = 0 $:

$$ \frac{4(4 - x^2)}{(x^2 + 4)^2} = 0 $$
$$ 4(4 - x^2) = 0 $$
$$ 4 - x^2 = 0 $$
$$ x^2 = 4 $$
$$ x = \pm 2 $$

So, the critical points are $ x = 2 $ and $ x = -2 $.

3. **Second Derivative (f''(x)):**

$$ f''(x) = \frac{d}{dx} \left( \frac{4(4 - x^2)}{(x^2 + 4)^2} \right) $$
$$ f''(x) = \frac{-8x(x^2 + 4)^2 - 2(4 - x^2)(2(x^2 + 4)(2x))}{(x^2 + 4)^4} $$
$$ f''(x) = \frac{-8x(x^2 + 4)^2 - 2(4 - x^2)(4x(x^2 + 4))}{(x^2 + 4)^4} $$

Now, evaluate at $ x = 2 $ and $ x = -2 $ to determine concavity and inflection points.

At $ x = 2 $:

$$ f''(2) = \frac{-8(2)(2^2 + 4)^2 - 2(4 - 2^2)(4(2)(2^2 + 4))}{(2^2 + 4)^4} $$
$$ f''(2) = \frac{-8(2)(36) - 2(0)(4(36))}{36^2} $$
$$ f''(2) = \frac{-576}{36^2} $$
$$ f''(2) = -\frac{16}{9} $$

At $ x = -2 $:

$$ f''(-2) = \frac{-8(-2)((-2)^2 + 4)^2 - 2(4 - (-2)^2)(4(-2)((-2)^2 + 4))}{((-2)^2 + 4)^4} $$
$$ f''(-2) = \frac{-8(-2)(36) - 2(0)(4(36))}{36^2} $$
$$ f''(-2) = \frac{576}{36^2} $$
$$ f''(-2) = \frac{16}{9} $$

Since $ f''(2) < 0 $ and $ f''(-2) > 0 $, we conclude that:

- At $ x = 2 $, the function is concave down, indicating a local maximum.
- At $ x = -2 $, the function is concave up, indicating a local minimum.

Therefore, $ f(x) = \frac{4x}{x^2 + 4} $ has a local maximum at $ x = 2 $ and a local minimum at $ x = -2 $. There are no inflection points.