What intervals is #f(x) = (5x^2)/(x^2 + 4)# concave up/down?

Question

Charles Arocha · Accepted Answer

In order to investigate concavity, we shall look at the sign of the second derivative.

#f(x) = (5x^2)/(x^2 + 4)#

#f'(x) = (-40x)/(x^2+4)^2#

Finding #f''(x)#:

#f''(x) = (-40(x^2+4)^2 -(-40x)2(x^2+4)(2x))/(x^2+4)^4#

#= ((x^2+4)[-40(x^2+4) + 160x^2])/(x^2+4)^4#

# = [-40(x^2+4) + 160x^2]/(x^2+4)^3 #

So:

#f''(x) = (120x^2 - 160)/(x^2+4)^3#

In general, a function can change sign by either crossing the #x# axis (being equal to #0#), or by being discontinuous (teleporting across the #x# axis).

In this case #f''(x)# is never discontinuous for real #x#, and

#f''(x) = 0# when #3x^2-4 = 0#, so #x= +-2/sqrt3#

Cut the number line into intervals and test each interval:

#(-oo, -2/sqrt3)# #f''(x)# is positive, so the graph of #f# is concave up

#(-2/sqrt3, 2/sqrt3)# #f''(x)# is negative, so the graph of #f# is concave down

#(2/sqrt3, oo)# #f''(x)# is positive, so the graph of #f# is concave up

Samantha Adkison · Answer

undefined To determine the intervals where the function $ f(x) = \frac{5x^2}{x^2 + 4} $ is concave up or concave down, we need to find the second derivative of the function and then analyze its sign.

Given $ f(x) = \frac{5x^2}{x^2 + 4} $, we first find the first derivative $ f'(x) $ using the quotient rule:

$$ f'(x) = \frac{(10x(x^2 + 4) - 5x^2(2x))}{(x^2 + 4)^2} $$

Simplifying:

$$ f'(x) = \frac{10x^3 + 40x - 10x^3}{(x^2 + 4)^2} $$
$$ f'(x) = \frac{40x}{(x^2 + 4)^2} $$

Now, to find the second derivative $ f''(x) $, we differentiate $ f'(x) $ with respect to $ x $:

$$ f''(x) = \frac{40(x^2 + 4)^2 - 80x^2(x^2 + 4)(2x)}{(x^2 + 4)^4} $$

Simplifying:

$$ f''(x) = \frac{40(x^4 + 8x^2 + 16) - 160x^3(x^2 + 4)}{(x^2 + 4)^3} $$
$$ f''(x) = \frac{40x^4 + 320x^2 + 640 - 160x^5 - 640x^3}{(x^2 + 4)^3} $$
$$ f''(x) = \frac{-160x^5 + 40x^4 - 640x^3 + 320x^2 + 640}{(x^2 + 4)^3} $$

Now, to determine the intervals where the function is concave up or concave down, we need to find the critical points of $ f''(x) $ and test the sign of $ f''(x) $ in those intervals.

However, since $ f''(x) $ is a polynomial, it has no critical points. Therefore, we'll need to analyze the sign of $ f''(x) $ by considering the behavior of $ x $ as it approaches positive and negative infinity.

As $ x $ approaches positive infinity, the dominant term in $ f''(x) $ is $ -160x^5 $, which is negative. Thus, $ f''(x) $ is negative for large positive $ x $ values, indicating that the function is concave down in that interval.

As $ x $ approaches negative infinity, the dominant term in $ f''(x) $ is also $ -160x^5 $, which is still negative. Thus, $ f''(x) $ is negative for large negative $ x $ values as well, indicating that the function is concave down in that interval.

Therefore, the function $ f(x) = \frac{5x^2}{x^2 + 4} $ is concave down for all real values of $ x $.