Maximum and minimum on a closed interval

For a differentiable function such as a polynomial, the maximum and minimum on a closed interval occur at an endpoint or at a point inside where $f'(x)=0$. There are three steps.

Solve $f'(x)=0$ and keep only the solutions inside the interval. Compute $f(x)$ at those points and at the endpoints $x=a,\ b$. The largest value is the maximum and the smallest is the minimum.

If $f'(x)=0$ has solutions $x=0,\ 4$ and the interval is $[1,\ 7]$, the $x$-values to check are $1,\ 4,\ 7$.

Examples

1. 01On 0 ≤ x ≤ 2, f'(x)=0 only at x=1. What is the maximum value?$\displaystyle f(0)=-1,\; f(1)=6,\; f(2)=-2$AnswerHide

   Answer$\displaystyle 6$

   Steps

   1. $\displaystyle 6$$\displaystyle > -1 > -2$
   2. The largest is f(1); the answer is that value
2. 02The x-values to check are 0, □, 5. Find □.$\displaystyle \begin{gathered} \displaystyle f'(x)=0 \;\Rightarrow\; x=3,\ 8, \\[6pt] \displaystyle [0,\ 5] \end{gathered}$AnswerHide

   Answer$\displaystyle 3$
3. 03To find the maximum and minimum on this interval, which x are the candidates?$\displaystyle \begin{gathered} \displaystyle f'(x)=0 \;\Rightarrow\; x=\pm3, \\[6pt] \displaystyle [0,\ 6] \end{gathered}$AnswerHide

   Answer$\displaystyle x=0,\ 3,\ 6$

   Steps

   1. Keep only the stationary points inside the interval
   2. $\displaystyle x$$\displaystyle = 3$
   3. x = -3 is outside the interval, so it does not count
   4. Then add the two endpoints
   5. $\displaystyle x$$\displaystyle = 0,\ 3,\ 6$
4. 04The x-values to check are 2, □, 7. Find □.$\displaystyle \begin{gathered} \displaystyle f'(x)=0 \;\Rightarrow\; x=5,\ 8, \\[6pt] \displaystyle [2,\ 7] \end{gathered}$AnswerHide

   Answer$\displaystyle 5$

   Steps

   1. Keep only the stationary points inside the interval
   2. $\displaystyle x$$\displaystyle = 5$
   3. x = 8 is outside the interval
   4. Adding the endpoints
   5. $\displaystyle x$$\displaystyle = 2,\ 5,\ 7$
5. 05The x-values to check are 1, □, 5. Find □.$\displaystyle \begin{gathered} \displaystyle f'(x)=0 \;\Rightarrow\; x=-1,\ 2, \\[6pt] \displaystyle [1,\ 5] \end{gathered}$AnswerHide

   Answer$\displaystyle 2$

   Steps

   1. Keep only the stationary points inside the interval
   2. $\displaystyle x$$\displaystyle = 2$
   3. x = -1 is outside the interval
   4. Adding the endpoints
   5. $\displaystyle x$$\displaystyle = 1,\ 2,\ 5$
6. 06The x-values to check are 1, □, 3. Find □.$\displaystyle \begin{gathered} \displaystyle f'(x)=0 \;\Rightarrow\; x=-2,\ 2, \\[6pt] \displaystyle [1,\ 3] \end{gathered}$AnswerHide

   Answer$\displaystyle 2$

   Steps

   1. Keep only the stationary points inside the interval
   2. $\displaystyle x$$\displaystyle = 2$
   3. x = -2 is outside the interval
   4. Adding the endpoints
   5. $\displaystyle x$$\displaystyle = 1,\ 2,\ 3$

FAQ

Q1Is a local maximum automatically the maximum?AnswerHide

A

Not necessarily. On $-2\le x\le2$, even if $f(-1)=-1$ is a local extremum, the endpoint value $f(2)=7$ is larger, so the maximum is $7$. Always compare the endpoint values too.

Q2What if a solution of $f'(x)=0$ lies outside the interval?AnswerHide

A

Drop it from the candidates. If there is no point inside where $f'(x)=0$, then $f(x)$ is monotonic on the interval, and the endpoints alone decide the maximum and minimum.