If # e^x / (5+e^x)#, what are the points of inflection, concavity and critical points?

Question

William Abdalla · Accepted Answer

The point of inflection is #(ln5,1/2)# and no critical points. The concavities are shown below

We need #(u/v)'=(u'v-uv')/(v^2)# Calculate the first and second derivatives Let #f(x)=e^x/(5+e^x)# #u(x)=e^x#, #=>#, #u'(x)=e^x# #v(x)=5+e^x#, #=>#, #v'(x)=e^x# Therefore, #f'(x)=(e^x(5+e^x)-e^x*e^x)/(5+e^x)^2=(5e^x)/(5+e^x)^2# #AA x in RR, |, f'(x)>0# No critical points. Therefore, The sign chart is #color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,+oo)# #color(white)(aaaa)##sign f'(x)##color(white)(aaaaaaaa)##+# #color(white)(aaaaaaa)## f(x)##color(white)(aaaaaaaaaa)##↗# Now, calculate the second derivative #u(x)=5e^x#, #=>#, #u'(x)=5e^x# #v(x)=(5+e^x)^2#, #=>#, #v'(x)=2e^x(5+e^x)# #f''(x)=(5e^x(5+e^x)^2-5e^x(2e^x(5+e^x)))/(5+e^x)^4# #=(25e^x+5e^(2x)-10e^(2x))/((5+e^x)^3)# #=(25e^x-5e^(2x))/(((5+e^x)^3))# #=(5e^x(5-e^(x)))/(((5+e^x)^3))# The point of inflection is when #f''(x)=0# #=>#, #5-e^x=0#, #x=ln5# We can make the chart #color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,ln5)##color(white)(aaaa)##(ln5,+oo)# #color(white)(aaaa)##sign f''(x)##color(white)(aaaaaa)##+##color(white)(aaaaaaaaaaa)##-# #color(white)(aaaa)## f(x)##color(white)(aaaaaaaaaaaa)##uu##color(white)(aaaaaaaaaaa)##nn# See the graph of the function graph{e^x/(5+e^x) [-8.89, 8.89, -4.444, 4.445]}

Elijah Abram · Answer

undefined To find the points of inflection, concavity, and critical points of the function $ \frac{e^x}{5+e^x} $, we first need to find its second derivative and analyze its behavior.

First, let's find the first derivative:

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

Simplify this to get:

$$ f'(x) = \frac{5e^x}{(5+e^x)^2} $$

Now, let's find the second derivative:

$$ f''(x) = \frac{d}{dx} \left(\frac{5e^x}{(5+e^x)^2}\right) $$

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

$$ f''(x) = \frac{5e^x(5+e^x - 10e^x)}{(5+e^x)^3} $$

$$ f''(x) = \frac{5e^x(5 - 9e^x)}{(5+e^x)^3} $$

The critical points occur where the first derivative equals zero or is undefined. Setting $ f'(x) = 0 $, we get:

$$ \frac{5e^x}{(5+e^x)^2} = 0 $$

Solving this equation yields no real solutions. So, there are no critical points.

The concavity of the function changes where the second derivative equals zero or is undefined. Setting $ f''(x) = 0 $, we get:

$$ 5 - 9e^x = 0 $$

Solving this equation yields:

$$ e^x = \frac{5}{9} $$

$$ x = \ln\left(\frac{5}{9}\right) $$

To determine the concavity at this point, evaluate $ f''(x) $ around $ x = \ln\left(\frac{5}{9}\right) $.

If $ f''(x) > 0 $, the function is concave up.
If $ f''(x) < 0 $, the function is concave down.

Substitute $ x = \ln\left(\frac{5}{9}\right) $ into $ f''(x) $ and determine its sign.

Since $ e^x > 0 $ for all real numbers $ x $, $ 5 - 9e^x $ is negative when $ x = \ln\left(\frac{5}{9}\right) $. Therefore, the function is concave down at this point.

As for points of inflection, since the function changes concavity at $ x = \ln\left(\frac{5}{9}\right) $, this point is also a point of inflection.

In summary:

- There are no critical points.
- The function is concave down at $ x = \ln\left(\frac{5}{9}\right) $.
- $ x = \ln\left(\frac{5}{9}\right) $ is a point of inflection.