Investigating logarithmic functions

Exponential and logarithmic functions, grade 12 (5 units) Lesson 15 of 23Learning

Investigating logarithmic functions

The domain of

\displaystyle \ln f(x)

$\displaystyle \ln f(x)$ is defined only where $\displaystyle f(x)>0$ . At an end of the domain where $\displaystyle f(x)$ comes close to 0, $\displaystyle \ln f(x)$ falls to $\displaystyle -\infty$ , so there is a vertical asymptote there.

For example, $\displaystyle {f(x)}={\ln(x^2-4x)}$ :

$\displaystyle x^2-4x>0$ where $\displaystyle x<0$ or $\displaystyle x>4$ – this is the domain;
as $\displaystyle x$ comes close to 0 from the left or to 4 from the right, $\displaystyle x^2-4x$ comes close to 0 and $\displaystyle f(x)$ falls to $\displaystyle -\infty$ ;
the vertical asymptotes: $\displaystyle {x}={0}$ and $\displaystyle {x}={4}$ .

\displaystyle f(x)=\ln(x^{2}-4x)

\displaystyle x=0

\displaystyle x=4

\displaystyle \ln x

grows more slowly than any power

$\displaystyle \ln x$ tends to infinity, but more slowly than any positive power of $\displaystyle x$ :

$\displaystyle x$	$\displaystyle 10$	$\displaystyle 100$	$\displaystyle 1000$	$\displaystyle 10^6$
$\displaystyle \frac{\ln x}{x}$	$\displaystyle 0.23$	$\displaystyle 0.046$	$\displaystyle 0.0069$	$\displaystyle 0.000014$

As $\displaystyle x\to\infty$ : $\displaystyle \frac{\ln x}{x}\to0$ , and so do $\displaystyle \frac{\ln x}{x^n}$ and $\displaystyle \frac{\ln x}{\sqrt{x}}$ .
As $\displaystyle x\to0^+$ : $\displaystyle x\ln x\to0$ – the factor $\displaystyle x$ “beats” $\displaystyle \ln x$ , which falls to $\displaystyle -\infty$ .

These facts give horizontal asymptotes and the behaviour of a graph at an end of its domain.

Investigating

\displaystyle \frac{\ln x}{x}

The example below investigates $\displaystyle {f(x)}={\frac{\ln x}{x}}$ , the second derivative included:

The domain: $\displaystyle x>0$ .
The axes: $\displaystyle {\ln x}={0}$ where $\displaystyle {x}={1}$ , that is $\displaystyle (1,0)$ . The graph does not meet the y-axis.
Asymptotes: as $\displaystyle x\to0^+$ , $\displaystyle f(x)\to-\infty$ – the vertical asymptote $\displaystyle {x}={0}$ ; as $\displaystyle x\to\infty$ , $\displaystyle f(x)\to0$ – the horizontal asymptote $\displaystyle {y}={0}$ .
$\displaystyle {f'(x)}={\frac{\frac{1}{x}\cdot x-\ln x}{x^2}}={\frac{1-\ln x}{x^2}}$ , and $\displaystyle {f'(x)}={0}$ where $\displaystyle {\ln x}={1}$ , that is $\displaystyle {x}={e}$ .
$\displaystyle f$ increases where $\displaystyle 0<x<e$ and decreases where $\displaystyle x>e$ : a maximum $\displaystyle \left(e,\frac{1}{e}\right)\approx(2.72,0.37)$ .
$\displaystyle {f''(x)}={\frac{-\frac{1}{x}\cdot x^2-(1-\ln x)\cdot2x}{x^4}}={\frac{2\ln x-3}{x^3}}$ , and $\displaystyle {f''(x)}={0}$ where $\displaystyle {\ln x}={\frac{3}{2}}$ , that is $\displaystyle {x}={e^{\frac{3}{2}}}$ .

\displaystyle f(x)=\frac{\ln x}{x}

\displaystyle x=0

\displaystyle M=(2.718,0.368)

\displaystyle P=(4.482,0.335)

Where $\displaystyle 0<x<e^{\frac{3}{2}}$ , $\displaystyle f''(x)<0$ – the function is concave down; where $\displaystyle x>e^{\frac{3}{2}}$ , $\displaystyle f''(x)>0$ – it is concave up.
An inflection point: $\displaystyle \left(e^{\frac{3}{2}},\frac{3}{2}e^{-\frac{3}{2}}\right)\approx(4.48,0.33)$ .

Investigating

\displaystyle x\ln x

The example below investigates $\displaystyle {f(x)}={x\ln x}$ :

The domain: $\displaystyle x>0$ . As $\displaystyle x\to0^+$ , $\displaystyle x\ln x\to0$ : the graph starts next to the origin, with no vertical asymptote.
The x-axis: $\displaystyle {x\ln x}={0}$ where $\displaystyle {x}={1}$ ( $\displaystyle {x}={0}$ is not in the domain).
$\displaystyle {f'(x)}={\ln x+1}$ , and $\displaystyle {f'(x)}={0}$ where $\displaystyle {\ln x}={-1}$ , that is $\displaystyle {x}={e^{-1}}$ .
$\displaystyle f$ decreases where $\displaystyle 0<x<\frac{1}{e}$ and increases where $\displaystyle x>\frac{1}{e}$ : a minimum $\displaystyle \left(\frac{1}{e},-\frac{1}{e}\right)\approx(0.37,-0.37)$ .
$\displaystyle {f''(x)}={\frac{1}{x}>0}$ throughout the domain: the function is concave up and has no inflection points.

\displaystyle f(x)=x\ln x

\displaystyle M=(0.368,-0.368)

An even function:

\displaystyle \ln(x^2+1)

The example below investigates $\displaystyle {f(x)}={\ln(x^2+1)}$ :

$\displaystyle x^2+1>0$ for every $\displaystyle x$ , so the domain is every $\displaystyle x$ and there is no vertical asymptote.
$\displaystyle {f(-x)}={\ln((-x)^2+1)}={f(x)}$ : the function is even, and its graph is symmetric about the y-axis.
$\displaystyle {f'(x)}={\frac{2x}{x^2+1}}$ : $\displaystyle f$ decreases where $\displaystyle x<0$ and increases where $\displaystyle x>0$ , a minimum $\displaystyle (0,0)$ .
$\displaystyle {f''(x)}={\frac{2(x^2+1)-2x\cdot2x}{(x^2+1)^2}}={\frac{2(1-x^2)}{(x^2+1)^2}}$ , and $\displaystyle {f''(x)}={0}$ where $\displaystyle {x}={\pm1}$ .
Concave up where $\displaystyle -1<x<1$ , concave down where $\displaystyle x<-1$ or $\displaystyle x>1$ .
Inflection points: $\displaystyle (-1,\ln2)$ and $\displaystyle (1,\ln2)$ .

\displaystyle f(x)=\ln(x^{2}+1)

\displaystyle P=(-1,0.693)

\displaystyle Q=(1,0.693)

Exercise 1

The function $\displaystyle {f(x)}={x-\ln x}$ is given.

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=x-\ln x$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {f'(x)}={0}$ .

Find $\displaystyle f''(x)$ .
Explain why the function has no inflection points, and find the type of the extremum point.

$\displaystyle f(x)=x-\ln x$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Mark the extremum point on the graph.

\displaystyle f(x)=x-\ln x

Exercise 2

The function $\displaystyle {f(x)}={\ln(x^2-4x+5)}$ is given.

Find the domain: solve the inequality.

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=\ln(x^{2}-4x+5)$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find $\displaystyle f''(x)$ , and simplify the numerator.

$\displaystyle f(x)=\ln(x^{2}-4x+5)$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {f''(x)}={0}$ .

Mark the extremum point and the inflection points on the graph.

\displaystyle f(x)=\ln(x^{2}-4x+5)

Exercise 3

The function $\displaystyle {f(x)}={(\ln x)^2}$ is given, for $\displaystyle x>0$ .

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=(\ln x)^{2}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find $\displaystyle f''(x)$ .

$\displaystyle f(x)=(\ln x)^{2}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find where the function is concave up: solve the inequality $\displaystyle f''(x)>0$ .

Mark the extremum point and the inflection point on the graph.

\displaystyle f(x)=(\ln x)^{2}

Exercise 4

The function $\displaystyle {f(x)}={\ln(x^2+a)}$ is given, with $\displaystyle a>0$ a parameter.
The function has an inflection point where $\displaystyle {x}={3}$ .

Find $\displaystyle f''(x)$ .

$\displaystyle f(x)=\ln(x^{2}+a)$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find $\displaystyle a$ .

Substitute the $\displaystyle a$ you found, and mark both inflection points on the graph.

\displaystyle f(x)=\ln(x^{2}+9)

Formulas: exponential and logarithmic functions, grade 12 (5 units)

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Powers

A product of powers with the same base
$\displaystyle a^{x}\cdot a^{y}$ $\displaystyle =$ $\displaystyle a^{x+y}$
A quotient of powers with the same base
$\displaystyle \frac{a^{x}}{a^{y}}$ $\displaystyle =$ $\displaystyle a^{x-y}$
A power of a power
$\displaystyle (a^{x})^{y}$ $\displaystyle =$ $\displaystyle a^{x\cdot y}$
A power of a product
$\displaystyle (a\cdot b)^{x}$ $\displaystyle =$ $\displaystyle a^{x}\cdot b^{x}$
A power of a quotient
$\displaystyle \left(\frac{a}{b}\right)^{x}$ $\displaystyle =$ $\displaystyle \frac{a^{x}}{b^{x}}$
A negative exponent
$\displaystyle a^{-x}$ $\displaystyle =$ $\displaystyle \frac{1}{a^{x}}$
The exponent 0
$\displaystyle a^{0}$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle a\neq0$
A rational exponent
$\displaystyle a^{\frac{m}{n}}$ $\displaystyle =$ $\displaystyle \sqrt[n]{a^{m}}$ $\displaystyle a>0$

Logarithms (within the domain)

The definition of the logarithm
$\displaystyle \log_{a}x=b\iff a^{b}=x$ $\displaystyle a>0,\ a\neq1,\ x>0$
$\displaystyle \log_{a}(a^{b})$ $\displaystyle =$ $\displaystyle b$
$\displaystyle a^{\log_{a}x}$ $\displaystyle =$ $\displaystyle x$
The logarithm of a product
$\displaystyle \log_{a}x+\log_{a}y$ $\displaystyle =$ $\displaystyle \log_{a}(x\cdot y)$
The logarithm of a quotient
$\displaystyle \log_{a}x-\log_{a}y$ $\displaystyle =$ $\displaystyle \log_{a}\left(\frac{x}{y}\right)$
The logarithm of a power
$\displaystyle \log_{a}(x^{b})$ $\displaystyle =$ $\displaystyle b\cdot\log_{a}x$
Change of base
$\displaystyle \log_{c}x$ $\displaystyle =$ $\displaystyle \frac{\log_{a}x}{\log_{a}c}$
$\displaystyle \log_{a}b$ $\displaystyle =$ $\displaystyle \frac{1}{\log_{b}a}$
The natural logarithm – to the base $\displaystyle e$
$\displaystyle \ln x=\log_{e}x$

The exponential and the logarithmic function

Every exponential as a power of $\displaystyle e$
$\displaystyle a^{x}$ $\displaystyle =$ $\displaystyle e^{x\ln a}$
$\displaystyle a^x$ and $\displaystyle \log_a x$ – inverse functions
$\displaystyle y=a^{x}\iff x=\log_{a}y$ $\displaystyle y>0$
The number $\displaystyle e$
$\displaystyle e\approx2.718$

Growth and decay

The amount after $\displaystyle t$ units of time
$\displaystyle f(t)=f(0)\cdot q^{t}$
The growth / decay factor – $\displaystyle p$ the percent per unit of time
$\displaystyle q=1\pm\frac{p}{100}$
The half-life $\displaystyle T$
$\displaystyle q^{T}$ $\displaystyle =$ $\displaystyle \frac{1}{2}$
The doubling time $\displaystyle T$
$\displaystyle q^{T}$ $\displaystyle =$ $\displaystyle 2$

Derivatives

$\displaystyle (e^{x})^{\prime}$ $\displaystyle =$ $\displaystyle e^{x}$
$\displaystyle (a^{x})^{\prime}$ $\displaystyle =$ $\displaystyle a^{x}\cdot\ln a$
$\displaystyle (\ln x)^{\prime}$ $\displaystyle =$ $\displaystyle \frac{1}{x}$
$\displaystyle (\log_{a}x)^{\prime}$ $\displaystyle =$ $\displaystyle \frac{1}{x\cdot\ln a}$
The derivative of $\displaystyle e^{f(x)}$
$\displaystyle \left[e^{f(x)}\right]^{\prime}$ $\displaystyle =$ $\displaystyle e^{f(x)}\cdot f^{\prime}(x)$
The derivative of $\displaystyle \ln(f(x))$
$\displaystyle \left[\ln(f(x))\right]^{\prime}$ $\displaystyle =$ $\displaystyle \frac{f^{\prime}(x)}{f(x)}$
The derivative of a product
$\displaystyle \left[f(x)\cdot g(x)\right]^{\prime}$ $\displaystyle =$ $\displaystyle f^{\prime}(x)\cdot g(x)+f(x)\cdot g^{\prime}(x)$
The derivative of a quotient
$\displaystyle \left[\frac{f(x)}{g(x)}\right]^{\prime}$ $\displaystyle =$ $\displaystyle \frac{f^{\prime}(x)\cdot g(x)-f(x)\cdot g^{\prime}(x)}{\left[g(x)\right]^{2}}$
The derivative of a composite function
$\displaystyle \left[f(u(x))\right]^{\prime}$ $\displaystyle =$ $\displaystyle f^{\prime}(u)\cdot u^{\prime}(x)$
The tangent’s equation at the point where $\displaystyle {x}={a}$
$\displaystyle y-f(a)=f^{\prime}(a)(x-a)$

Integrals, the area and accumulation

$\displaystyle \int e^{x}\,dx=e^{x}+C$
$\displaystyle \int a^{x}\,dx=\frac{a^{x}}{\ln a}+C$
In any interval where $\displaystyle x\neq0$
$\displaystyle \int\frac{1}{x}\,dx=\ln|x|+C$
$\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int f(mx+b)\,dx=\frac{1}{m}F(mx+b)+C$ $\displaystyle m\neq0$
$\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int f(u(x))\cdot u^{\prime}(x)\,dx=F(u(x))+C$
Where $\displaystyle f(x)\neq0$
$\displaystyle \int\frac{f^{\prime}(x)}{f(x)}\,dx=\ln|f(x)|+C$
$\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int_{a}^{b}f(x)\,dx=F(b)-F(a)$
The area between two graphs
$\displaystyle S=\int_{a}^{b}\left(f(x)-g(x)\right)dx$ $\displaystyle f(x)\geq g(x)$
The accumulation function’s derivative
$\displaystyle \left(\int_{a}^{x}f(t)\,dt\right)^{\prime}=f(x)$