- 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 \(y=x\).

- 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: \[\begin{equation}\tag{*} \log_b(x)=y\text{ if and only if }x=b^y\text{ for $x>0$.} \end{equation}\]
- 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

- Once again, the logarithmic function \(\log_b(x)\) is defined by \[\begin{equation}\tag{*} b^y=x\text{ if and only if }y=\log_b(x)\text{ for $x>0$.} \end{equation}\]
*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.

- 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.

- 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: \(\arccos(x)\), \(\arcsin(x)\), \(\arctan(x)\).
- Click here for graphs.
- Exercises: 1.5: 64, 90, 70