Find absolute extrema of polynomial with unknown power coefficients

Question

Find absolute extrema of the function $$f(x) = x^a(1-x)^b$$ on $[0,1]$ where $a,b>0$.

Answer 1

We can find the absolute extrema on a closed interval by the Closed Interval Method:

1. Find the values of $f$ at the critical numbers of $f$ in $(a,b)$.
2. Find the values of $f$ at the endpoints of the interval.
3. The largest of the values from $(1)$ and $(2)$ is the absolute maximum; the smallest of these values is the absolute minimum.

Step 1:

Observe that because $a,b>0$ the derivative will exist (if either of them was negative, we could have a denominator that was different than one).

We need to find the critical numbers in $(0,1)$ by finding $f'(x)$ and setting $f'(x)=0$. We have that $f(x)= x^a(1-x)^b$. Therefore

\begin{align}f'(x)&=\frac{d}{dx}\Big(x^a(1-x)^b\Big)\\&= \frac{d}{dx}\big(x^a\big)(1-x)^b+x^a\frac{d}{dx}\big((1-x)^b\big)\\&= ax^{a-1}(1-x)^b+x^a\big(-b(1-x)^{b-1}\big)\\&= ax^{a-1}(1-x)^b-bx^a(1-x)^{b-1}\\&= x^{a-1}(1-x)^{b-1}\big(a(1-x)-bx\big)\\&= x^{a-1}(1-x)^{b-1}\big(a-(a+b)x\big) \end{align}

where upon setting $f'(x)=0$ we have that $a-(a+b)x=0$ so that

$$x=\frac{a}{a+b}$$

is our critical number (the other two terms produce $x=1$ provided $b>1$ and $x=0$ provided $a>1$ which are covered by testing our end points). In most calculus courses these critical points are denoted using $c$ instead of $x$. Therefore we have that our critical number is

$$c=\frac{a}{a+b}$$

We need to find the value of $f$ at this critical number. We have that

$$f\Big(\frac{a}{a+b}\Big)=\Big(\frac{a}{a+b}\Big)^a\Big(1-\Big(\frac{a}{a+b}\Big)\Big)^b$$

which I will let you simplify with the hint that

$$1=\frac{a+b}{a+b}$$

Step 2:

Evaluate $f(0)$ and $f(1)$.

Step 3:

Compare the values of $f(c),f(0),$ and $f(1)$. The largest of these values is the absolute maximum while the smallest is the absolute minimum.