How do you sketch the graph #y=(2e^x)/(1+e^(2x))# using the first and second derivatives?

Answer 1

Luke Alvarez

First we note that as #e^(2x) > 0# the denominator of the quotient never vanishes, so that:

#f(x) = (2e^x)/(1+e^(2x))#

is defined and continuous for all #x in RR#.

By writing the function as:

#f(x) = 2/((1+e^(2x))/e^x) = 2/(e^-x+e^x) = 1/cosh x#

we can also see that the function is even and at the limits of the domain of definition we have:

#lim_(x->-oo) 2/(e^-x+e^x) = lim_(x->oo) 2/(e^-x+e^x)= 0#

So that #f(x)# has #y=0# as an horizontal asymptote on both sides.

Evaluate now the first and second derivatives:

#(df)/dx = d/dx (1/cosh) = -sinhx/cosh^2x#

#(d^2f)/dx^2 = -d/dx (sinhx/cosh^2x) = -(cosh^3x - 2sinh^2xcoshx)/cosh^4x = (2sinh^2x-cosh^2x)/cosh^3x = (sinh^2x-1)/cosh^3x#

We have then that #(df)/dx > 0# for #x < 0# and #(df)/dx < 0# for #x > 0# while the only stationary point where #(df)/dx = 0# is for #x=0#.

The function is therefore strictly increasing in #(-oo,0)#, strictly decreasing in #(0,+oo)# and reaches a maximum for #x=0# ov value #f(0) = 1#. As #coshx >=1# we can see that this is an absolute maximum.

The second derivative is null for #sinhx = +-1#, that is for:

#x = ln(sqrt2+-1)#

These are two inflection points and the function is concave down in the interval #(ln(sqrt2-1),ln(sqrt2+1)) #and concave up outside this interval.

graph{ (2e^x)/(1+e^(2x)) [-3.59, 3.59, -1.794, 1.796]}

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Answer 2

Cameron Alexander

To sketch the graph $y = \frac{2e^x}{1 + e^{2x}}$ using the first and second derivatives:

Find the first derivative $y'$ : $y' = \frac{d}{dx} \left( \frac{2e^x}{1 + e^{2x}} \right)$ $y' = \frac{2e^x(1 + e^{2x}) - 2e^x(2xe^{2x})}{(1 + e^{2x})^2}$ $y' = \frac{2e^x + 2e^{3x} - 4xe^{3x}}{(1 + e^{2x})^2}$
Find the critical points by setting $y' = 0$ : $2e^x + 2e^{3x} - 4xe^{3x} = 0$
Solve for $x$ to find critical points.
Determine the intervals where $y' > 0$ and $y' < 0$ to identify increasing and decreasing segments.
Find the second derivative $y''$ : $y'' = \frac{d}{dx} \left( \frac{2e^x}{1 + e^{2x}} \right)'$
Evaluate $y''$ at the critical points found earlier to determine concavity.
Plot the critical points, increasing/decreasing intervals, and concavity to sketch the graph.
Determine the behavior of the function as $x$ approaches positive and negative infinity to complete the sketch.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.