3.1.11 Function Inverses

Usually on the last column you are guaranteed to have to compute the inverse of a function at a particular value. The easiest way to do this is to not explicitly solve for the inverse and plug in the point but rather, compute the inverse at that point as you go. For example if you are given a function f(x)=32x−2f(x) = \frac{3}{2}x - 2 and you want to calculate f−1(x)f^{-1}(x) at the point x=3x = 3, you don’t want to do the standard procedure for finding inverses (switch the xx and yy variables and solve for yy) which would be:

x=32y−2⇒y=(x+2)⋅23x = \frac{3}{2}y - 2 \Rightarrow y = (x + 2) \cdot \frac{2}{3}

at x=3x=3: ⇒y=(3+2)⋅23=103\Rightarrow y = (3 + 2) \cdot \frac{2}{3} = \frac{10}{3}

Not only do you solve for the function, you have to remember the function while you’re plugging in numbers. An easier way is just switch the xx and yy variables, then plug in the value for xx, then compute yy. That way you aren’t solving for the inverse function for all points, but rather the inverse at that particular point. Let’s see how doing that procedure would look like:

x=32y−2⇒3=32y−2⇒y=(3+2)⋅23=103x = \frac{3}{2}y - 2 \Rightarrow 3 = \frac{3}{2}y - 2 \Rightarrow y = (3 + 2) \cdot \frac{2}{3} = \frac{10}{3}

Although this might not seem like much, it does help in saving some time.

Another important thing to remember when computing inverses is a special case when the function is in the form:

f(x)=ax+bcx+d⇒f−1(x)=−dx+bcx−af(x) = \frac{ax + b}{cx + d} \Rightarrow f^{-1}(x) = \frac{-dx + b}{cx - a}

This was a very popular trick awhile back, but slowly it’s appearance has been dwindling, however that does not mean a resurgence is unlikely. The important thing to remember is to line up the xx’s on the numerator and denominator so it is in the require form. Here is an example problem to show you the trick:

Example: Find f−1(2)f^{-1}(2) where f(x)=2x+34+5xf(x) = \frac{2x + 3}{4 + 5x}. Solution: f(x)=2x+34+5x=2x+35x+4⇒f−1(x)=−4x+35x−2⇒f−1(2)=−4⋅2+35⋅2−2=−58f(x) = \frac{2x + 3}{4 + 5x} = \frac{2x + 3}{5x + 4} \Rightarrow f^{-1}(x) = \frac{-4x + 3}{5x - 2} \Rightarrow f^{-1}(2) = \frac{-4 \cdot 2 + 3}{5 \cdot 2 - 2} = \frac{-5}{8}

Here are some problems to give you some practice:

Problem Set 3.1.11

f(x)=3x+2,f−1(−2)=f(x) = 3x + 2, f^{-1}(-2) =
f(x)=4x5,f−1(2)=f(x) = \frac{4x}{5}, f^{-1}(2) =
f(x)=2−3x,f−1(1)=f(x) = 2 - 3x, f^{-1}(1) =
f(x)=x2−1f(x) = x^2 - 1 and x>0,f−1(8)=x > 0, f^{-1}(8) =
f(x)=5+3x,f−1(−2)=f(x) = 5 + 3x, f^{-1}(-2) =
f(x)=4−3x,f−1(2)=f(x) = 4 - 3x, f^{-1}(2) =
f(x)=83+x,f−1(2)=f(x) = \frac{8}{3 + x}, f^{-1}(2) =
f(x)=3−2x4,f−1(−1)=f(x) = \frac{3 - 2x}{4}, f^{-1}(-1) =
f(x)=x33+3,f−1(−6)=f(x) = \frac{x^3}{3} + 3, f^{-1}(-6) =
f(x)=2−3x4,f−1(5)=f(x) = 2 - \frac{3x}{4}, f^{-1}(5) =
f(x)=2x+1,f−1(3)=f(x) = 2x + 1, f^{-1}(3) =
g(x)=3x+2,g−1(−1)=g(x) = 3x + 2, g^{-1}(-1) =
h(x)=2x−3,h−1(−1)=h(x) = 2x - 3, h^{-1}(-1) =
f(x)=2(x+3),f−1(−4)=f(x) = 2(x + 3), f^{-1}(-4) =
f(x)=2−3x,f−1(4)=f(x) = 2 - 3x, f^{-1}(4) =
h(x)=5x−3,h−1(2)=h(x) = 5x - 3, h^{-1}(2) =
h(x)=5−3x,h−1(−2)=h(x) = 5 - 3x, h^{-1}(-2) =
f(x)=2x+2,f−1(−2)=f(x) = 2x + 2, f^{-1}(-2) =
f(x)=3x−3,f−1(−3)=f(x) = 3x - 3, f^{-1}(-3) =
f(x)=4−4x,f−1(−4)=f(x) = 4 - 4x, f^{-1}(-4) =
f(x)=3x−1x−3,f−1(1)=f(x) = \frac{3x - 1}{x - 3}, f^{-1}(1) =
f(x)=2x+1x−2,f−1(3)=f(x) = \frac{2x + 1}{x - 2}, f^{-1}(3) =
f(x)=3x−1x−3,f−1(−1)=f(x) = \frac{3x - 1}{x - 3}, f^{-1}(-1) =
f(x)=1−3xx+3,f−1(−2)=f(x) = \frac{1 - 3x}{x + 3}, f^{-1}(-2) =