Derivatives provide a large amount of information about the shape of a the graph of a function. First, we will consider whether functions are increasing or decreasing, that is, if they are going up or down. Recall the definitions of a function being increasing or decreasing on an interval.
Definition4.1.1.
A function \(f\) is increasing on an interval \((a,b)\) if for all \(x_1, x_2 \in (a,b)\text{,}\) if \(x_1 \lt x_2\text{,}\) then \(f(x_1) \lt f(x_2)\text{.}\)
A function \(f\) is decreasing on an interval \((a,b)\) if for all \(x_1, x_2 \in (a,b)\text{,}\) if \(x_1 \gt x_2\text{,}\) then \(f(x_1) \gt f(x_2)\text{.}\)
Subsection4.1.1Increasing/Decreasing Test (Positive Derivative Implies Increasing)
Consider a differentiable function \(y = f(x)\text{.}\) Recall that the value of the first derivative \(f'\) represents the slope of the tangent line of \(f\text{.}\) If \(f'(x) > 0\text{,}\) then a tangent line has positive slope, and so the graph is sloping upwards and so is increasing. Similarly, if \(f'(x) \lt 0\text{,}\) then the tangent line has negative slope and \(f\) is decreasing.
Show that \(e^x > 1 + x\) for \(x > 0\text{.}\) Then, from that, show that \(e^x > 1 + x + \frac{x^2}{2}\) for \(x > 0\text{.}\) By induction, show that \(e^x > 1 + x + \frac{x^2}{2} + \dots + \frac{x^n}{n!}\) for \(x > 0\text{,}\) for any \(n\text{.}\)
Subsection4.1.3Increasing Functions and Inverse Functions
Recall that functions which are increasing (or decreasing) on an interval are one-to-one on that interval, and thus have an inverse function.
Theorem4.1.4.
If \(f'(x) > 0\) for all \(x \in (a,b)\) (or, \(f'(x) \lt 0\)), then \(f\) is one-to-one on \((a,b)\)
Consider the cubic function \(f(x) = x^3 + 4x + 7\text{.}\) We know that some cubic functions are one-to-one, but some have 2 turning points and are not one-to-one. In this case, \(f'(x) = 3x^2 + 4 > 0\) for all \(x\text{,}\) so \(f\) is one-to-one.
Subsection4.1.4Increasing Functions and Roots
Recall that by the intermediate value theorem, if \(f(a) \lt 0\) and \(f(b) > 0\text{,}\) then there exists at least one solution \(c \in (a,b)\) such that \(f(c) = 0\text{.}\) Using the fact that positive derivative implies increasing, we can additionally show that there is at most one solution in the interval, which together implies that there is exactly one solution in the interval.
If \(f'(x) > 0\) on \((a,b)\text{,}\) then there is at most one root. This is because if there were two roots \(c_1\) and \(c_2\text{,}\) say with \(c_1 \lt c_2\text{,}\) with \(f(c_1) = f(c_2) = 0\text{.}\) However, beacuse the function is increasing, \(c_1 \lt c_2\) implies that \(f(c_1) \lt f(c_2)\text{,}\) which contradicts the fact that they are both solutions.
Theorem4.1.6.
If \(f'(x) \neq 0\) on an interval, then \(f\) has at most one root in that interval.
