Short and Sweet Calculus

4.1 Increasing and Decreasing Functions

  • A function is increasing if as we move along the curve from left to right the curve is rising (Figure [fig:Ch6-Increasing-Fig1](a)). Mathematically speaking, a function \(f\) is increasing on an interval \(I\) if \[f(x_{1})<f(x_{2})\quad\text{whenever}\quad x_{1}<x_{2}\text{ in }I\]

  • A function is decreasing if as we move along the curve from left to right, the curve is falling (Figure [fig:Ch6-Increasing-Fig1](b)). Mathemtically speaking, \(f\) is a decreasing function on \(I\) if \[f(x_{1})>f(x_{2})\quad\text{whenever}\quad x_{1}<x_{2}\text{ in }I.\]

  • A function is constant if its graph is horizontal (Figure [fig:Ch6-Increasing-Fig1](c)). Mathematically speaking, we say \(f\) is constant on \(I\) if \[f(x_{1})=f(x_{2})\quad\text{for all points }x_{1}\text{ and }x_{2}\text{ in }I.\]

    (a) Increasing function. \(x_{1}<x_{2}\Rightarrow f(x_{1})<f(x_{2})\). (b) Decreasing function. \(x_{1}<x_{2}\Rightarrow f(x_{1})<f(x_{2})\). (c) Constant function. For every \(x_{1}\) and \(x_{2}\) in the domain: \(f(x_{1})=f(x_{2})\).

A function may be sometimes increasing and sometimes decreasing (or neither). For example, the graph of \(f(x)=2x^{3}-12x^{2}+18x-2\) is illustrated in Figure 4.1. At \(x=1\) the function ceases to increase and begins decreasing; at \(x=3\), the reverse is true. At \(x=1\) and \(x=3\) the tangent to the curve is evidently parallel to the \(x\)-axis, meaning the slope is zero. We will talk about this kind of points in Section [sec:Ch4-Extreme-values].

Graph of f(x)=x^{3}-12x^{2}+18x-2.

Slope of graph and increasing and decreasing functions

It is evident from Figure 4.2 that on the interval where a function \(f(x)\) is increasing, the tangent line makes a positive acute angle \(\theta\) with the positive \(x\)-axis; hence \[\text{slope}=\tan\theta=f'(x)>0\] and on the interval where the function is decreasing, the tangent line makes an obtuse angle \(\theta\) with the positive \(x\)-axis; hence \[\text{slope}=\tan\theta=f'(x)<0.\]

4.1.0.1 Tests for determining where a function is increasing and where decreasing

From the above observations, we may guess that \(f\) is increasing function if \(f'(x)>0\) and is decreasing if \(f'(x)<0\). Let’s see how we can verify that.

First let us suppose that \(f'(x)=dy/dx\) is positive. Then since \(dy/dx\) is the limit of \(\Delta y/\Delta x\), it follows that, we have \[\frac{\Delta y}{\Delta x}>0,\] for sufficiently small values of \(\Delta x\). This means that if \(\Delta x=x-x_{0}\) is positive then \(\Delta y=f(x)-f(x_{0})\) is also positive; that is, \[\Delta x=x-x_{0}>0\Rightarrow\Delta y=f(x)-f(x_{0})>0\] or \[x_{0}<x\Rightarrow f(x_{0})<f(x).\] Therefore, an increase of \(x\) causes an increase of \(y=f(x)\). From this argument, we learn that we can verify a differentiable function is increasing where its derivative is positive \[f'(x)=\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}>0.\] Analogously, if \(dy/dx<0\), it follows that \(\Delta y/\Delta x<0\) for sufficiently small values of \(\Delta x\); that is if \(\Delta x>0\), then \(\Delta y<0\) or an increase of \(x\) causes a decrease of \(y=f(x)\).

4.1. Increasing/Decreasing Test.

(a) If \(f'(x)>0\) for every \(x\) in an interval, then \(f\) is increasing \((\nearrow)\) on that interval

(b) If \(f'(x)<0\) for every \(x\) in an interval, then \(f\) is decreasing \((\searrow)\) on that interval.

(c) If \(f'(x)=0\) for every \(x\) in an interval, then \(f\) is constant on that interval.

In applying this theorem, we need to determine the sign of \(f'(x)\). If \(f'(x)\) is a polynomial, this may be conveniently done by breaking it up into factors and determining the sign of each factor. A factor of the form \(x-a\) is negative when \(x<a\) and is positive when \(x>a\).

Suppose, then, we wish to determine the sign of \[(x+2)(x-1)(x-3).\]

  1. When \(-\infty<x<-2\), all three factors are negative, and thus the product is negative \[(-)(-)(-)=(-).\]

  2. When \(-2<x<1\), the first factor is positive, and the other factors are negative. Therefore, the product is positive \[(+)(-)(-)=(+).\]

  3. When \(1<x<3\), the first two factors are positive, but the last factor is negative. Therefore, the product is negative \[(+)(+)(-)=(-).\]

  4. When \(3<x<+\infty\), all three factors are positive, and the product is positive too \[(+)(+)(+)=(+).\]

Obviously the product is zero at \(-2,1\), and \(3\). We may summarize these into the following “sign table.”

\(x\) \(-\infty\) \(-2\) 1 3 \(+\infty\)
sign of \((x+2)\) \(-\) \(0\) \(+\) \(+\) \(+\) \(+\) \(+\)
sign of \((x-1)\) \(-\) \(-\) \(-\) \(0\) \(+\) \(+\) \(+\)
sign of \((x-3)\) \(-\) \(-\) \(-\) \(-\) \(-\) \(0\) \(+\)
\(\therefore\) sign of product \(-\) \(0\) \(+\) \(0\) \(-\) \(0\) \(+\)

Example 4.1. Determine where \(f(x)=x^{2}e^{-x}\) is increasing and where it is decreasing.

Solution

Let’s find the derivative of \(f\): \[f(x)=x^{2}e^{-x}\Rightarrow f'(x)=2xe^{-x}-x^{2}e^{-x}=e^{-x}(2x-x^{2}).\] Because \(e^{-x}>0\), the sign of \(f'(x)\) is solely determined by \(2x-x^{2}=x(2-x).\)

\(x\) \(-\infty\) 0 2 \(+\infty\)
sign of \(x\) \(—\) \(0\) \(+++\) \(+\) \(+++\)
sign of \(2-x\) \(+++\) \(+\) \(+++\) \(0\) \(—\)
\(\therefore\) sign of \(f'(x)\) \(—\) \(0\) \(+++\) \(0\) \(—\)
Increaing/Decreasing \(f(x)\) \(\searrow\) \(\nearrow\) \(\searrow\)

This table shows that \(f(x)\) is decreasing on \((-\infty,0]\cup[2,+\infty)\) (because \(f'(x)<0\)) and is increasing on \([0,2]\) (because \(f'(x)>0\)). The graph of \(f\) is shown in the following figure.


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