exponential functions

[startingover]

f(x) = e^x + e^-x
f'(x) = e^x (1) + e^-x(-1)
f'(x) = e^x - e^-x

f"(x) = e^x (-1) +e^x (1)
f"(x) = -e^x + e^x

How do you solve to find the inflection points and concavity?

 

[galactus]

The second derivative of \(\displaystyle \L\\e^{x}+e^{-x}\) is the same,

\(\displaystyle \L\\e^{x}+e^{-x}=0\)

This does not have any real solutions which resolves to 0.

See?. No inflection points.

Since f''(x)>0 on the interval \(\displaystyle \L\\({-\infty},{\infty})\), then it is concave up.

 

[soroban]

Hello, startingover!

\(\displaystyle f(x) \:= \:e^x\,+\,e^{-x }\)

\(\displaystyle f'(x)\:=\:e^x\,-\,e^{-x}\)

\(\displaystyle f''x)\:=\:e^x\,+\,e^{-x }\)

How do you solve to find the inflection points and concavity?

As usual, solve: \(\displaystyle f''(x)\:=\:0\)

We have: \(\displaystyle \:e^x\,+\,e^{-x}\:=\:0\)

Multiply by \(\displaystyle e^x:\;\;e^{2x}\,+\,1\:=\:0\;\;\Rightarrow\;\;e^{2x}\:=\:-1\)

Take logs: \(\displaystyle \:\ln\left(e^{2x}\right) \:=\:\ln(-1)\) ? . . . no real roots

Therefore, there are no inflection points.


At \(\displaystyle x\,=\,0:\;\;f''(0) \:=\:e^0\,+\,e^0 \:=\:+2\) . . . always positive.

Therefore, the graph is always concave up
. . as shown in galactus' graph.