1. points of inflection

Determine the points of inflection in each graph.

a) f(x)=xe^(-x)
f''(x)=e^(-x)(e^(-x) +x)
f''(x)=0 DNE
therefore no poitns of inflection

b) f(x)= e^x + e^(-x)
f''(x)=e^x+ e^(-x)
f''(x)= 0 DNE
therefore no points of inflection

c) f(x)= x-lnx
f''(x)=(x^2+1)/x^2
f''(x)=undefined when x=0
therefore point of inflection is (0,0)

d) f(x)=x^4 + 4x^3
f''(x)=12x^2+24x
f''(x)=0 when x= -2 or x=0
therefore points of inflection are (0,0) and (-2,-16)

let me know which ones are wrong and the correct method to answering the question.

2. Rethink the first one.
$\begin{gathered}
f(x) = xe^{ - x} \hfill \\
f'(x) = e^{ - x} - xe^{ - x} = e^{ - x} \left( {1 - x} \right) \hfill \\
\end{gathered}$

You need to learn basic algebra.

3. Originally Posted by skeske1234
Determine the points of inflection in each graph.

a) f(x)=xe^(-x)
f''(x)=e^(-x)(e^(-x) +x) ... f''(x) is incorrect
f''(x)=0 DNE
therefore no poitns of inflection ... an inflection point exists at x = 2

b) f(x)= e^x + e^(-x)
f''(x)=e^x+ e^(-x)
f''(x)= 0 DNE
therefore no points of inflection

c) f(x)= x-lnx
f''(x)=(x^2+1)/x^2 f''(x) = 1/x^2
f''(x)=undefined when x=0
therefore point of inflection is (0,0) the function itself is undefined at x = 0 ... no inflection point.

d) f(x)=x^4 + 4x^3
f''(x)=12x^2+24x
f''(x)=0 when x= -2 or x=0
therefore points of inflection are (0,0) and (-2,-16)
...

4. Hello, skeske1234!

Determine the points of inflection for each function.

$a)\;\; f(x)\:=\:xe^{-x}$
$f'(x) \;=\;-xe^{-x} + e^{-x}$

$f''(x) \;=\;xe^{-x} - e^{-xc} - e^{-x} \;=\;xe^{-x} - 2e^{-x}$

$f''(x) \:=\:0 \quad\Rightarrow\quad e^{-x}(x-2) \:=\:0 \quad\Rightarrow\quad x = 2$

Then: . $f(2) \:=\:2e^{-2}$

. . Inflection point: . $\left(2,\:\frac{2}{e^2}\right)$

$b)\;\;f(x)\:=\: e^x + e^{-x}$
$f'(x) \;=\;e^x - e^{-x}$

$f''(x)\:=\:e^x + e^{-x} \:=\:0$

Multiply by $e^x\!:\quad e^{2x} + 1 \:=\:0 \quad\Rightarrow\quad e^{2x} \:=\:-1 \;\;\hdots\;\text{DNE}$

. . Therefore, no inflection point.

$c)\;\; f(x)\:=\: x-\ln x$
$f'(x) \:=\:1 - \frac{1}{x} \:=\:1 - x^{-1}$

$f''(x) \:=\:x^{-2} \:=\:0 \quad\Rightarrow\quad \frac{1}{x^2} \:=\:0 \;\;\hdots\text{ DNE}$

. . Therefore, no inflection point.

Part (d) is correct!

5. Ok, I have a question regarding part c > the rational function. How come my teacher said that to find the points of inflection you're supposed to find the values of x when f''(x)=0 and f''(x)=undefined? When we find what value makes f''(x)=undefined, isnt that also a point of inflection? If it isn't, what is the reason behind finding the values of x that make f''(x)=undefined? and same thing with finding critical numbers, my teacher says to look for the values of x that make f'(x)=0 and f'(x)=undefined... Is this true here too? Why are we looking for the undefined derivative and what does it help us determine relating to points of inflection and or critical points?

Now back to part c) how come no points of inflection exist when shown from soroban's answer, he gets 0 and 0 is an answer isn't it? why isn't the point of inflection (0,0)? Because for this question, there is one i did in class relating to this example as: find the inflection points of f(x)=x^(1/3)
for this example, i'm reading here... it says:
f''(x)=-2/(9x^(5/3))
set f''(x)= 0
f''(x)= DNE when x=0
Point of inflection (0,0)

So what is the difference between these two questions..? If it says f''(x)= undefined at x=0 in the denominator, so point of inflection is (0,0)

6. If a curve has a point of inflection at x = c , then f''(c) is either equal to 0 or is undefined.

Note that the converse of this statement is false ... if f''(c) = 0 or is undefined, that does not mean an inflection point exists at x = c.

We use critical values of f'' (f'' equal to 0 or undefined) to help locate possible locations for points of inflection based on the true statement above ... that does not mean you will always find them there.

f(x) = x - ln(x) has f''(0) as undefined ... but f(0) itself is undefined ... how can a point of inflection exist there?

finally and most importantly, for an inflection point to exist on a curve at x = c ...
1. f(x) must be continuous at x = c.
2. the concavity of f(x) changes at x = c, i.e. f''(x) must change sign at x = c.