Find the points at which $f$ has an absolute maximum or minimum on $I$ without graphing

Question

Assume $I=[0.9,3.1], f:I\rightarrow\mathbb{R}$ is defined by $f(x):=|x^2-4x+3|, x\in I$. Without sketching the graph of $f$ on $I$, find points at which $f$ has an absolute minimum on $I$ and points at which $f$ has an absolute minimum on $I$.

We know that $f(0.9)=0.21$ and $f(3.1)=0.21$. I am a bit confused on how I can find the absolute minimum/maximum, however, without being able to differentiate since $f$ is not differentiable on $I$.

I know that obviously the max occurs at $x=2$ and the mins occur at $x=1,3$, but I am not sure how to show this without graphing.

Answer 1

Since $x^2-4x+3=(x-1)(x-3),$ the absolute value will be nondifferentiable at $1$ and $3$, each of which is in your interval $[0.9,3.1]$ and are therefore critical numbers to be tested. [each will give $f(x)=0$] Then the absolute value will be $+(x^2-4x+3)$ with derivative $2x-4$ outside of $[1,3]$ and this could contributes a new critical number at 2 but 2 is not outside of $[1,3]$. Similarly the absolute value inside $(1,3)$ is $-(x^2-4x+3)$ with derivative $-2x+4$ again possibly making the critical number 2, but this time 2 indeed is inside $(1,3)$ so is a critical number to be tested. At this point you have the endpoints $0.9,\ 3.1$ along with the three internal critical numbers at $1,2,3$ and all these should be put into $f(x)$ to get the absolute max and min.

Edit:Initially I erroneously threw out $2$ as a critical number above. I re-typed things to include that critical number.

Answer 2

For the minimum, it is easy to see that it must be greater or equal zero. Thus, you can first assume it to be zero and solve the equation $ x^2-4x+3=0 $. Either this equation has a real solution in your interval (in this case you are done), or not. In the latter case you would know that the curve does not cross $ y=0 $, and since $ y (2)> 0 $ and since the curve is continuous, you could look on the curve without absolute value and solve it normally.