One of the most basic and essential results in differential calculus known as the “first derivative theorem” states that the derivative of a function at an extremum in an open interval inside the domain of the function vanishes. Note that in the post on the extreme value theorem, we showed that a continuous real-valued function defined over a closed interval attains its minimum and maximum at some points in the closed interval. However, the extreme value theorem does not tell us exactly where these extremes occur. The first derivative theorem states that if we want to find the extremes of differentiable functions, we need to see where the derivatives of these functions vanish.

#### Definition of local (relative) and global (absolute) extremum

By definition, a function \(ƒ\) has a local maximum (minimum) value at a point \(c \in D_f\) if $$f(x) \leq f(c) (f(x) \geq f(c))$$ for all \(x\) in some open interval \(I \subseteq D_f\) containing \(c\).

A function has a local (also called relative) extremum if it has either a local maximum or a local minimum.

The largest value of a function on a set is its global (absolute) maximum and the smallest value of a function is its global (absolute) minimum. By a global (absolute) extremum, we mean either a global maximum or a global minimum.

**Example**. The function \(f(x) = \sin \displaystyle \frac{1}{x} \) has infinitely many local maximums and local minimums in every interval of the form \((0,a)\) where \(a\) is a positive real number.

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

**Result**. Let \(f: (a,b) \rightarrow \mathbb{R}\) be a function reaching its maximum or minimum at a point \(c \in (a,b)\). If \(f\) is differentiable at \(c\), then \(f'(c) = 0\).

The proof goes as follows:

Let \(f\) reach to its maximum at \(c\). Our claim is that \(f'(c) = 0\). Suppose not. Then, either \(f'(c) > 0\) or \(f'(c) < 0\).

For the moment, let \(f'(c) > 0\). This means that $$\lim_{h \to 0} \frac{f(c+h) – f(c)}{h} > 0$$ which implies that for small enough \(h\), $$\frac{f(c+h) – f(c)}{h}$$ is positive. We may choose \(h\) positive and so small that \(c+h\) is in the open interval \((a,b)\). Therefore, \(f(c+h) > f(c)\) which is a contradiction because \(f\) is supposed to reach to its maximum at \(c\).

Similarly, if we assume that \(f'(c) < 0\) and \(h\) is negative and small enough, we will have that \(f(c+h) > f(c)\). This is again a contradiction.

A similar argument shows that if \(f\) reaches to its minimum at \(c\), then \(f'(c) = 0\) and the proof is complete.

#### Some problems for the first derivative theorem

**Exercise**. Find the local extremums of the function $$f(x) = 2x^3+3x^2 – 12x +1$$ defined on the interval \([-3,3]\).

**Solution**. First we consider the function \(f\) over the open interval \((-3,3)\). Since the function \(f\) is differentiable everywhere, by the first derivative theorem, we need to solve the equation $$f'(x) = 6x^2 + 6x – 12 = 0$$ to find the extremums of the function \(f\). It is easy to see that if \(f'(x) = 0\), then either \(x = -2\) or \(x = 1\). Now, observe that $$f(-2) = 21 \text{ and } f(1) = -6.$$ On the other hand, $$f(-3) = 10 \text{ and } f(3) = 46.$$

Therefore, \(f\) over \([-3,3]\) has the global minimum at \(1\) and the global maximum at \(3\). However, \(f\) has a local maximum at \(-2\).

**Exercise**. Find the absolute extremums of the function \(g(x) = \sqrt[3]{x^4}\) on the interval \([-1, 8]\).

**Solution**. The equation \(g'(x)=0\) has only one root and that is \(x=0\). Note that \(g(0) = 0\).

Also, observe that $$g(-1) = 1 \text{ and } g(8) = 16.$$ So, \(g\) has a global minimum at \(0\) and global maximum at \(8\).