The first derivative test gives a criterion to identify points in the domain of a function at which it has extreme values. In this post, we state and prove the first derivative test and we use this test to find extreme values of some functions.

#### A statement and a proof of the first derivative test

**Result** (the first derivative test). Let \(f\) be continuous on an interval containing a point \(c\). Then, the following statements hold:

- If \(f'(x)\) is positive for all \(x\) less than \(c\) and negative for all \(x\) more than \(c\), then \(f\) attains its maximum on the interval at \(c\).
- If \(f'(x)\) is negative for all \(x\) less than \(c\) and positive for all \(x\) more than \(c\), then \(f\) attains its minimum on the interval at \(c\).

For the proof of the first derivative test use the criteria for monotonicity discussed in the post on the mean value theorem.

#### Finding extreme values with the help of the first derivative test

**Exercise**. Let \(a\), \(b\), and \(c\) be real numbers such that \(a > 0\). Find the extreme values of the following polynomial function: $$q(x) = ax^2 + bx + c.$$

**Solution**. Observe that \(q'(x) = 2ax+b.\) Since \(a > 0\), we have that \(q’\) is negative if \(x\) is less than \(-b/2a\) and \(q’\) is positive if \(x\) is more than \(-b/2a\). Therefore, by the first derivative test, \(q\) attains its minimum at \(-b/2a\). Note that $$q(x) = \frac{(2ax+b)^2}{4a} + \frac{4ac-b^2}{4a}$$ and $$q(-b/2a) = \frac{4ac-b^2}{4a}.$$

**Exercise**. Find the extreme values of the function \(g(x) = x – \sin x\) over the interval \([0,\pi]\).

**Solution**. Observe that \(g'(x) = 1 – \cos x\) which is always non-negative. Therefore, \(g\) is increasing, and so, over the interval \([0,\pi]\), it attains its minimum at \(0\) and its maximum at \(\pi\).

**Exercise**. Find the extreme values of the function \(h(x) = x – \ln x\) over its domain.

**Exercise**. Find the extreme values of the sinc function defined as follows: $$\hbox{sinc}(x)= \begin{cases} \displaystyle \frac{\sin x}{x} & \text{ if } x \neq 0 \\ 1 & \text{ if } x = 0 \end{cases}.$$