We must show that [latex]{f}^{-1}\left(f\left(x\right)\right)=x[/latex] and [latex]f\left({f}^{-1}\left(x\right)\right)=x[/latex].

[latex]\begin{align}{f}^{-1}\left(f\left(x\right)\right)&={f}^{-1}\left(\frac{1}{x+1}\right) \\[1mm] &=\dfrac{1}{\frac{1}{x+1}}-1 \\[1mm] &=\left(x+1\right)-1 \\[1mm] &=x \\[5mm] f\left({f}^{-1}\left(x\right)\right)&=f\left(\frac{1}{x}-1\right) \\[1mm] &=\dfrac{1}{\left(\frac{1}{x}-1\right)+1} \\[1mm] &=\dfrac{1}{\frac{1}{x}} \\[1mm] &=x \end{align}[/latex]

Therefore, [latex]f\left(x\right)=\dfrac{1}{x+1}[/latex] and [latex]{f}^{-1}\left(x\right)=\dfrac{1}{x}-1[/latex] are inverses.

This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for [latex]x[/latex].

[latex]\begin{align}y&=5{x}^{3}+1 \\[1mm] x&=5{y}^{3}+1 \\[1mm] x - 1&=5{y}^{3} \\[1mm] \dfrac{x - 1}{5}&={y}^{3} \\[4mm] {f}^{-1}\left(x\right)&=\sqrt[3]{\dfrac{x - 1}{5}} \end{align}[/latex]

Analysis of the Solution

Look at the graph of [latex]f[/latex] and [latex]{f}^{-1}[/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[/latex]. This is always the case when graphing a function and its inverse function.

Also, since the method involved interchanging [latex]x[/latex] and [latex]y[/latex], notice corresponding points. If [latex]\left(a,b\right)[/latex] is on the graph of [latex]f[/latex], then [latex]\left(b,a\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Since [latex]\left(0,1\right)[/latex] is on the graph of [latex]f[/latex], then [latex]\left(1,0\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Similarly, since [latex]\left(1,6\right)[/latex] is on the graph of [latex]f[/latex], then [latex]\left(6,1\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex].

The original function [latex]f\left(x\right)={\left(x - 4\right)}^{2}[/latex] is not one-to-one, but the function is restricted to a domain of [latex]x\ge 4[/latex] or [latex]x\le 4[/latex] on which it is one-to-one.

To find the inverse, start by replacing [latex]f\left(x\right)[/latex] with the simple variable [latex]y[/latex].

[latex]\begin{align}y&={\left(x - 4\right)}^{2} &&\text{Interchange } x \text{ and }y. \\[1mm] x&={\left(y - 4\right)}^{2} &&\text{Take the square root}. \\[1mm] \pm \sqrt{x}&=y - 4 &&\text{Add } 4 \text{ to both sides}. \\[1mm] 4\pm \sqrt{x}&=y \end{align}[/latex]

This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of [latex]x[/latex] and [latex]y[/latex] for the original [latex]f(x)[/latex], we looked at the domain: the values [latex]x[/latex] could assume. When we reversed the roles of [latex]x[/latex] and [latex]y[/latex], this gave us the values [latex]y[/latex] could assume. For this function, [latex]x\ge 4[/latex], so for the inverse, we should have [latex]y\ge 4[/latex], which is what our inverse function gives.

1. The domain of the original function was restricted to [latex]x\ge 4[/latex], so the outputs of the inverse need to be the same, [latex]f\left(x\right)\ge 4[/latex], and we must use the + case:
   [latex]{f}^{-1}\left(x\right)=4+\sqrt{x}[/latex]
2. The domain of the original function was restricted to [latex]x\le 4[/latex], so the outputs of the inverse need to be the same, [latex]f\left(x\right)\le 4[/latex], and we must use the – case:
   [latex]{f}^{-1}\left(x\right)=4-\sqrt{x}[/latex]

Analysis of the Solution

On the graphs below, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line [latex]y=x[/latex]. The coordinate pair [latex]\left(4, 0\right)[/latex] is on the graph of [latex]f[/latex] and the coordinate pair [latex]\left(0, 4\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. For any coordinate pair, if [latex](a,b)[/latex] is on the graph of [latex]f[/latex], then [latex](b,a)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Finally, observe that the graph of [latex]f[/latex] intersects the graph of [latex]{f}^{-1}[/latex] on the line [latex]y=x[/latex]. Points of intersection for the graphs of [latex]f[/latex] and [latex]{f}^{-1}[/latex] will always lie on the line [latex]y=x[/latex].

We can see this is a parabola with vertex at [latex]\left(2, -3\right)[/latex] that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to [latex]x\ge 2[/latex].

To find the inverse, we will use the vertex form of the quadratic. We start by replacing [latex]f(x)[/latex] with a simple variable, [latex]y[/latex], then solve for [latex]x[/latex].

[latex]\begin{align}y&={\left(x - 2\right)}^{2}-3 && \text{Interchange } x \text{ and } y. \\[1mm] x&={\left(y - 2\right)}^{2}-3 && \text{Add 3 to both sides}. \\[1mm] x+3&={\left(y - 2\right)}^{2} && \text{Take the square root}. \\[1mm] \pm \sqrt{x+3}&=y - 2 && \text{Add 2 to both sides}. \\[1mm] 2\pm \sqrt{x+3}&=y && \text{Rename the function}. \\[3mm] {f}^{-1}\left(x\right)&=2\pm \sqrt{x+3} \end{align}[/latex]

Now we need to determine which case to use. Because we restricted our original function to a domain of [latex]x\ge 2[/latex], the outputs of the inverse should be the same, telling us to utilize the + case

[latex]{f}^{-1}\left(x\right)=2+\sqrt{x+3}[/latex]

If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.

Analysis of the Solution

Notice that we arbitrarily decided to restrict the domain on [latex]x\ge 2[/latex]. We could just have easily opted to restrict the domain on [latex]x\le 2[/latex], in which case [latex]{f}^{-1}\left(x\right)=2-\sqrt{x+3}[/latex]. Observe the original function graphed on the same set of axes as its inverse function in the graph below. Notice that both graphs show symmetry about the line [latex]y=x[/latex]. The coordinate pair [latex]\left(2,\text{ }-3\right)[/latex] is on the graph of [latex]f[/latex] and the coordinate pair [latex]\left(-3,\text{ }2\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Observe from the graph of both functions on the same set of axes that

[latex]\text{domain of }f=\text{range of } {f}^{-1}=\left[2,\infty \right)[/latex]

and

[latex]\text{domain of }{f}^{-1}=\text{range of } f=\left[-3,\infty \right)[/latex]

Finally, observe that the graph of [latex]f[/latex] intersects the graph of [latex]{f}^{-1}[/latex] along the line [latex]y=x[/latex].

Note that the original function has range [latex]f\left(x\right)\ge 0[/latex]. Replace [latex]f\left(x\right)[/latex] with [latex]y[/latex], then solve for [latex]x[/latex].

[latex]\begin{align} y&=\sqrt{x - 4} && \text{Replace}f\left(x\right)\text{with }y.\\[1mm] x& =\sqrt{y - 4}&& \text{Interchange }x\text{ and }y. \\[1mm] x& =\sqrt{y - 4}&& \text{Square each side}. \\[1mm] {x}^{2}& =y - 4&& \text{Add 4}. \\[1mm] {x}^{2}+4&=y&& \text{Rename the function }{f}^{-1}\left(x\right). \\[4mm] {f}^{-1}\left(x\right)& ={x}^{2}+4 \end{align}[/latex]

Recall that the domain of this function must be limited to the range of the original function.

[latex]{f}^{-1}\left(x\right)={x}^{2}+4,x\ge 0[/latex]

Analysis of the Solution

Notice in the graph below that the inverse is a reflection of the original function over the line [latex]y=x[/latex]. Because the original function has only positive outputs, the inverse function has only positive inputs.

Start with the given function for [latex]V[/latex]. Notice that the meaningful domain for the function is [latex]r\ge 0[/latex] since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is [latex]V\ge 0[/latex]. Solve for [latex]r[/latex] in terms of [latex]V[/latex], using the method outlined previously.

[latex]\begin{align}V&=\frac{2}{3}\pi {r}^{3} \\[1mm] {r}^{3}&=\dfrac{3V}{2\pi } && \text{Solve for }{r}^{3}. \\[1mm] r&=\sqrt[3]{\frac{3V}{2\pi }} && \text{Solve for }r. \end{align}[/latex]

This is the result stated in the section opener. Now evaluate this for [latex]V=100[/latex] and [latex]\pi =3.14[/latex].

[latex]\begin{align}r&=\sqrt[3]{\dfrac{3V}{2\pi }} \\[1mm] &=\sqrt[3]{\dfrac{3\cdot 100}{2\cdot 3.14}} \\[1mm] &\approx \sqrt[3]{47.7707} \\ & \approx 3.63 \end{align}[/latex]

Therefore, the radius is about 3.63 ft.

Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where [latex]\dfrac{\left(x+2\right)\left(x - 3\right)}{\left(x - 1\right)}\ge 0[/latex]. The output of a rational function can change signs (change from positive to negative or vice versa) at [latex]x[/latex]-intercepts and at vertical asymptotes. For this equation the graph could change signs at [latex]x=[/latex] –2, 1, and 3.

To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph.

This function has two [latex]x[/latex]-intercepts, both of which exhibit linear behavior near the [latex]x[/latex]-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a [latex]y[/latex]-intercept at (0, 6).

From the [latex]y[/latex]-intercept and [latex]x[/latex]-intercept at [latex]x=–2[/latex], we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.

From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function [latex]f(x)[/latex] will be defined. [latex]f(x)[/latex] has domain [latex]-2\le x<1[/latex] or [latex]x\ge 3[/latex], or in interval notation, [latex]\left[-2,1\right)\cup \left[3,\infty \right)[/latex].

We first want the inverse of the function. We will solve for [latex]n[/latex] in terms of [latex]C[/latex].

[latex]\begin{align}C&=\dfrac{20+0.4n}{100+n}\\[1mm] C\left(100+n\right)&=20+0.4n\\[1mm] 100C+Cn&=20+0.4n\\[1mm] 100C - 20&=0.4n-Cn\\[1mm] 100C - 20&=\left(0.4-C\right)n\\[3mm] n&=\dfrac{100C - 20}{0.4-C}\end{align}[/latex]

Now evaluate this function for [latex]C=[/latex]0.35 (35%).

[latex]\begin{align}n&=\dfrac{100\left(0.35\right)-20}{0.4 - 0.35}\\[1mm] &=\frac{15}{0.05}\\[1mm] &=300\end{align}[/latex]

We can conclude that 300 mL of the 40% solution should be added.