# From last time

• The last thing we did on Wednesday was to find the inverses of the functions $$f(x)=2x-3$$, $$x^3+2$$, and $$x^2-x$$, the latter with domain $$x\ge\frac12$$.
• We said that you can do this by solving the equation $$y=f(x)$$ for $$x$$.
• In case of $$f(x)=x^2-x$$, we started out by solving $$x^2-x=x^2+2ax$$ for $$a$$.
• This is because in general we have $$x^2+2ax=(x+a)^2-a^2$$.
• Therefore, in the case of $$f(x)=x^2-x$$, we've got $$a=-\frac12$$, and therefore $$x^2-x=(x-\frac12)^2+\frac14$$.
• Click here for the graphs of the functions and their inverses.
• Also, note that you can get a value $$f(a)$$ by finding the intersection of the graph of $$f(x)$$ with the vertical line $$x=a$$.
• Similarly, to get a value $$f^{-1}(b)$$, you can find the intersection of the graph of $$f(x)$$ with the horizontal line $$y=b$$.
• Therefore, the relabeling $$x\leftrightarrow y$$ to get an inverse function $$f^{-1}(x)$$ with variable $$x$$ corresponds to relabeling the axes on the graph.
• Correspondingly the graph of the inverse $$y=f^{-1}(x)$$ is obtained by mirroring the graph of the original function $$y=f(x)$$ with respect to the line $$y=x$$.

# Definition of logarithmic functions

• Let $$b>0,\,b\ne1$$ be a number. Then the exponential function $$f(x)=b^x$$ is strictly monotonous.
• Since $$f(x)=b^x$$ is strictly monotonous, it is one-to-one, and therefore it has an inverse.
• The inverse $$f^{-1}(x)=\log_b(x)$$ is called the logarithmic function with base $$b$$.
• Since the exponential function $$f(x)=b^x$$ has domain $$(-\infty,\infty)$$ and range $$(0,\infty)$$, the logarithmic function $$f^{-1}(x)=\log_b(x)$$ has domain $$(0,\infty)$$ and range $$(-\infty,\infty)$$.
• It is defined by the inverse function definition: $$$\tag{*} \log_b(x)=y\text{ if and only if }x=b^y\text{ for x>0.}$$$
• The cancellation equations become the following formulas. \begin{align*} \log_b(b^x)=x&\text{ for all x},\\ b^{\log_b(x)}=x&\text{ for x>0}. \end{align*}
• The logarithmic function with base $$e$$ is called the natural logarithm function, and it is denoted by $$\ln(x)=\log_e(x)$$.
• Exercises: 1.5: 36, 52

# Laws of logarithms and Base change formula

• Once again, the logarithmic function $$\log_b(x)$$ is defined by $$$\tag{*} b^y=x\text{ if and only if }y=\log_b(x)\text{ for x>0.}$$$
• Laws of logarithms. We will get these formulas from the laws of exponents and the defining equations (*), via the substitution $$u=\log_b(x)$$ and $$v=\log_b(y)$$.
• Note that by cancellation we have $$b^u=b^{\log_b(x)}=x$$ and $$b^v=b^{\log_b(y)}=y$$.
• $$b^{u+v}=b^ub^v\leadsto u+v=\log_b(b^ub^v)\leadsto\log_b(x)+\log_b(y)=\log_b(xy)$$
• $$b^{u-v}=b^u/b^v\leadsto u-v=\log_b(b^u/b^v)\leadsto\log_b(x)-\log_b(y)=\log_b(x/y)$$
• $$b^{ru}=(b^u)^r\leadsto ru=\log_b((b^u)^r)\leadsto r\log_b(x)=\log_b(x^r)$$
• Base change formula. Let us pick an additional number $$a>0$$, $$a\ne1$$.
• Putting $$y=\log_b(x)$$, then by (*), we get $$b^y=x$$.
• Applying $$\log_a(x)$$ to both sides gives $$\log_a(b^y)=\log_a(x)$$.
• The third Law of logarithms gives $$y\log_a(b)=\log_a(x)$$.
• Substituting $$y=\log_b(x)$$ and rearranging gives the following formula. $\log_b(x)=\frac{\log_a(x)}{\log_a(b)}.$
• Exercises: 1.5: 40, 52, 54.

# Graphs and properties of logarithmic functions

• Notice the following properties of the logarithmic functions $$f(x)=\log_b(x)$$ on their graphs.
• If $$b>1$$, then the function is strictly monotonously increasing.
• If $$b<1$$, then the function is strictly monotonously decreasing.
• Whatever $$b\ne1$$ is, the function goes through $$(1,0)$$.
• Note that the rate of growth of the logarithmic functions is very slow.
• Exercise. 1.5: 56.

# Inverse trigonometric functions

• Although the trigonometric functions are periodic, and thus very much not one-to-one, we can restrict their domains to define their inverse functions.
• Recall the graph of the function $$f(x)=\cos(x)$$. We want to choose a domain on which it goes through its entire range $$[-1,1]$$.
• A standard choice is $$[0,\pi]$$. Thus, we define the inverse cosine function, with domain $$[-1,1]$$ as follows. $\cos^{-1}(x)=y\text{ if and only if }\cos(y)=x\text{ and }0\le y\le\pi$
• Similarly, we need to find an interval on which $$\sin(x)$$ goes through its entire range $$[-1,1]$$.
• A standard choice is $$[-\frac{\pi}{2},\frac{\pi}{2}]$$. Thus, we define the inverse sine function, with domain $$[-1,1]$$ as follows. $\sin^{-1}(x)=y\text{ if and only if }\sin(y)=x\text{ and }-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}$
• Finally, we need to find an interval on which $$\tan(x)$$ goes through its entire range $$(-\infty,\infty)$$.
• A standard choice is $$(-\frac{\pi}{2},\frac{\pi}{2})$$. Thus, we define the inverse tangent function, with domain $$(-\infty,\infty)$$ as follows. $\tan^{-1}(x)=y\text{ if and only if }\tan(y)=x\text{ and }-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}$
• Alternative notation: $$\arccos(x)$$, $$\arcsin(x)$$, $$\arctan(x)$$.