The maximum value of f(x)=x^50-x^20 in the interval [ 0, 1], is

The correct answer is Option (3) → 0

On a closed interval, the absolute maximum is the largest of the values at the critical points and at the end points.

    \( \displaystyle f'(x)=50x^{49}-20x^{19} \)

    \( \displaystyle =10x^{19}(5x^{30}-2) \)

f′(x) = 0 inside (0, 1) when

    \( \displaystyle x^{30}=\frac{2}{5} \)

At this point \( x^{20}=\left(x^{30}\right)^{2/3}=\left(\frac{2}{5}\right)^{2/3} \), so

    \( \displaystyle f=x^{20}\left(x^{30}-1\right) \)

    \( \displaystyle =\left(\frac{2}{5}\right)^{2/3}\left(\frac{2}{5}-1\right) \)

    \( \displaystyle =-\frac{3}{5}\left(\frac{2}{5}\right)^{2/3} \)

At the end points:

    \( \displaystyle f(0)=0 \)
    \( \displaystyle f(1)=1-1=0 \)

The values are \( -\frac{3}{5}\left(\frac{2}{5}\right)^{2/3} \) (negative) and 0, so the maximum value is 0 (and the critical point gives the minimum).

Shortcut: For \( 0\le x\le 1 \), a higher power is never larger: \( x^{50}\le x^{20} \). So \( f(x)\le 0 \) on [0, 1], and f(0) = 0. The maximum value is 0.

Tip: Option (1) is the value at the only critical point, but that point is a minimum. On a closed interval, compare with the values at the end points before calling a critical value the maximum.