Lower and Upper Bound calculation

Question

I'm working 'Integration' from Shurman's book and I'm trying to solve a problem.Here it is:

Let $J = [0, 1]$. Compute $mJ (f )$ and $MJ (f )$ for each of the following functions $$\begin{array}{rcl} f : J &\longrightarrow& R\\ x&\longmapsto& f (x) = x(1 − x)\end{array}$$

Solution:

It was given that :

• $mJ (f ) = \inf \{f (x) : x ∈ J \}$
• $MJ (f ) = \sup \{f (x) : x ∈ J \}$

So that I took the first derivative and found that $x=1/2$ is a critical point and $f(1/2)=1/4$ and $f(0)=f(1)=0$ so using that can I directly say that $mJ(f)=0$ and $MJ(f)=1/2$ ?

Thanks in advance!

Answer 1

Yes, you're right, but let me show you how to use algebra-precalculus and symmetry to solve this

• for fun, and
• for simplicity

We just need to have the shape of $f$ in mind. I write out the explanation since this is an answer. You may simplify the arguments in tests.

Observe that $J=[0,1]$ and $$f(x)=x(1-x)=\left(x+\frac12-\frac12\right)\left(\frac12+\frac12-x\right)=\left(\frac12\right)^2-\left(\frac12-x\right)^2$$ are symmetric about the line $x=1/2$.

Moreover, $f$ is a parabola with vertex $(1/2,1/4)$ with negative coefficient for $x^2$, so when $x\to\infty$, the term $-x^2$ dominates other terms.

Hence $m=f(0)=0$, and $M=f(1/2)=1/4$.

Answer 2

Note that $$ f(x)=x\cdot (1-x)=-x^2+x=-x^2+2\cdot\frac{1}{2}\cdot x-\frac{1}{4}+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4} $$