2. Images and preimages · Real functions

Image of a value

Definition

The image of a value $x_0 \in \operatorname{Dom} f$ is the value $f(x_0)$, that is, the value of $y$ associated with $x_0$. Since $f$ is a function, this value is unique.

\operatorname{Im}(x_0) = f(x_0).

Computing an image is always direct: just substitute.

Example 1 — Image from a formula: $f(x) = \log_{3}(x^{2}+1)$

What is the image of $2\sqrt{2}$? Substitute:

f(2\sqrt{2}) = \log_{3}\!\bigl((2\sqrt{2})^{2} + 1\bigr) = \log_{3}(8+1) = \log_{3} 9 = 2.

\boxed{\;f(2\sqrt{2}) = 2.\;}

Example 2 — Image from a table of values

A function is given by the table:

$x$	$-3$	$-2$	$-1$	$0$	$1$
$f(x)$	$3\sqrt{3}$	$3\sqrt{2}$	$3$	$0$	$\tfrac{1}{3}$

What is the image of $-2$? We read the table:

\boxed{\;f(-2) = 3\sqrt{2}.\;}

As before: the image is unique — in the column for $-2$ there is only one value.

Example 3 — Image from a graph

Given the following graph, we read off the images of $0$ and $-2$ by drawing a vertical line from the value of $x$ up to the curve and, from there, a horizontal line towards the $y$-axis:

We read:

f(0) = 2 \qquad\text{and}\qquad f(-2) = 0.

Preimage of a value

Definition

The preimage of a value $y$ is the set of all values of $x$ such that $f(x) = y$. It is written:

f^{-1}(y) = \{\,x \in \operatorname{Dom} f \;\mid\; f(x) = y\,\}.

Unlike the image, a preimage may have none, one or several solutions: you must solve the equation $f(x) = y$.

Strategy

To compute $f^{-1}(y)$:

1) Set up the equation $\;f(x) = y\;$ with $y$ as a specific value.

2) Solve it — using all the tools from the previous unit (polynomial, rational, exponential, logarithmic, irrational, etc.).

3) Check that the solutions found are in the domain of $f$. Those that are not are discarded.

Example — Preimages of $4$ for $f(x) = x^{2} - 5$

We set up $f(x) = 4$ and solve:

x^{2} - 5 = 4 \;\Longrightarrow\; x^{2} = 9 \;\Longrightarrow\; x = \pm\sqrt{9} = \pm 3.

Both solutions are real and lie in the domain ($\operatorname{Dom} f = \mathbb{R}$). Therefore:

\boxed{\;f^{-1}(4) = \{-3,\; +3\}.\;}

Geometrically, intersect the parabola $y = x^{2} - 5$ with the horizontal line $y=4$: it meets it at two points, $x=-3$ and $x=3$.

Image vs preimage — the essential difference

Image: evaluate — always a single result.

Preimage: solve an equation — 0, 1 or more results.

For example, with $f(x) = x^{2} - 5$:

· $\;f^{-1}(-5) = \{0\}$ (one preimage: $x^{2}=0$).

· $\;f^{-1}(4) = \{-3,\,3\}$ (two preimages).

· $\;f^{-1}(-10) = \varnothing$ (no preimage: $x^{2} = -5$ has no real solution).

Application — Efficiency of a company

The model: $P(t) = \dfrac{100}{1 + 4\,e^{-0{,}05\,t}}$

A company's efficiency percentage as a function of the days $t$ elapsed since it started is modelled by the function above. We will use it to see the three types of calculation: domain, image and preimage.

1) Domain. It is a rational function with denominator $1 + 4e^{-0{,}05t}$. We set it to zero to see whether it rules out any value:

1 + 4e^{-0{,}05t} = 0 \;\Longrightarrow\; 4e^{-0{,}05t} = -1 \;\Longrightarrow\; e^{-0{,}05t} = -\tfrac{1}{4}.

But an exponential is never negative, so it has no solution — the denominator never vanishes. Moreover, physically $t$ represents a number of days $\geq 0$, so:

\boxed{\;\operatorname{Dom} P = [0,\,+\infty).\;}

2) Images — initial efficiency and at 20 days

What is the initial efficiency? We compute $P(0)$:

P(0) = \dfrac{100}{1 + 4 e^{-0{,}05 \cdot 0}} = \dfrac{100}{1 + 4 \cdot 1} = \dfrac{100}{5} = 20.

\boxed{\;P(0) = 20\,\%.\;}

And after 20 days? Now $P(20)$:

P(20) = \dfrac{100}{1 + 4\,e^{-0{,}05 \cdot 20}} = \dfrac{100}{1 + 4\,e^{-1}} \approx \dfrac{100}{1 + 1{,}4715} \approx 40{,}46.

\boxed{\;P(20) \approx 40{,}46\,\%.\;}

Both calculations are direct images — substitute and compute.

3) Preimage — on which day is $70\,\%$ efficiency reached?

Now the question is inverse: we look for $t$ such that $P(t) = 70$. We set up the equation:

70 = \dfrac{100}{1 + 4\,e^{-0{,}05\,t}}.

Multiply by the denominator and isolate the exponential:

70\bigl(1 + 4 e^{-0{,}05 t}\bigr) = 100 \;\Longrightarrow\; 70 + 280\,e^{-0{,}05 t} = 100 \;\Longrightarrow\; 280\,e^{-0{,}05 t} = 30.

e^{-0{,}05 t} = \dfrac{30}{280} = \dfrac{3}{28}.

Apply the natural logarithm to both sides to "remove" the exponential:

-0{,}05\,t = \ln\!\dfrac{3}{28} \;\Longrightarrow\; t = -\dfrac{1}{0{,}05}\,\ln\!\dfrac{3}{28} \approx 44{,}67.

Since $t$ represents a whole number of days, we round up — $70\,\%$ efficiency is first reached at the end of day 45:

\boxed{\;t \approx 45\;\text{days}.\;}

The preimage is unique because $P(t)$ is strictly increasing: for each value of $y \in (20, 100)$ there is exactly one $t \geq 0$.

Exercises

1 Computing images

For each function, compute the requested image. Remember: just substitute and operate.

a) If $f(x) = 2x - 5$, compute $f(3)$.

Substitute $x=3$:

f(3) = 2 \cdot 3 - 5 = 6 - 5 = 1.

\boxed{\;f(3) = 1.\;}

b) If $f(x) = x^{2} + 1$, compute $f(-2)$.

$f(-2) = (-2)^{2} + 1 = 4 + 1 = 5$. Be careful to put the $-2$ in brackets before squaring.

\boxed{\;f(-2) = 5.\;}

c) If $f(x) = \log_{2}(x+4)$, compute $f(4)$.

$f(4) = \log_{2}(4+4) = \log_{2} 8 = 3$ (because $2^{3} = 8$).

\boxed{\;f(4) = 3.\;}

d) If $f(x) = \dfrac{6}{x-1}$, compute $f(4)$.

$f(4) = \dfrac{6}{4-1} = \dfrac{6}{3} = 2$. Check that $4 \in \operatorname{Dom} f = \mathbb{R} \setminus \{1\}$ ✓.

\boxed{\;f(4) = 2.\;}

2 Computing preimages

Compute all preimages of the requested value. Remember: there may be none, one or several — and you must check that the solutions are in the domain.

a) If $f(x) = 3x + 1$, find $f^{-1}(7)$.

We set up $3x + 1 = 7$ and isolate $x$:

3x = 6 \;\Longrightarrow\; x = 2.

A single preimage (affine functions are always injective).

\boxed{\;f^{-1}(7) = \{2\}.\;}

b) If $f(x) = x^{2} - 1$, find $f^{-1}(8)$.

$x^{2} - 1 = 8 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3$. Both solutions are in the domain $\mathbb{R}$.

\boxed{\;f^{-1}(8) = \{-3,\;3\}.\;}

c) If $f(x) = x^{2} + 4$, find $f^{-1}(0)$.

$x^{2} + 4 = 0 \Rightarrow x^{2} = -4$, which has no real solutions.

\boxed{\;f^{-1}(0) = \varnothing\;\;\text{(no preimage).}\;}

Geometrically: the parabola $y = x^{2} + 4$ has its vertex at $(0,4)$ and never touches the line $y=0$.

d) If $f(x) = \log(x-1)$, find $f^{-1}(2)$.

We set up $\log(x-1) = 2$. By the inverse definition of the common (base-10) logarithm ($\log = \log_{10}$):

x - 1 = 10^{2} = 100 \;\Longrightarrow\; x = 101.

Verification: $x = 101$ is in the domain ($x - 1 = 100 > 0$) ✓.

\boxed{\;f^{-1}(2) = \{101\}.\;}

3 Application — depreciation of a car

The value of a car (in thousands of euros) as a function of the years $t$ since purchase is modelled by $V(t) = 24 \cdot 0{,}85^{t}$.

a) What is the value of the car at purchase ($t=0$)?

It is an image: $V(0) = 24 \cdot 0{,}85^{0} = 24 \cdot 1 = 24$.

\boxed{\;V(0) = 24\,000\;\text{€}.\;}

b) What is the value after 5 years?

$V(5) = 24 \cdot 0{,}85^{5} \approx 24 \cdot 0{,}4437 \approx 10{,}65$.

\boxed{\;V(5) \approx 10\,650\;\text{€}.\;}

c) After how many years will the car be worth half ($12$ thousand €)?

It is a preimage: we look for $t$ with $V(t) = 12$.

24 \cdot 0{,}85^{t} = 12 \;\Longrightarrow\; 0{,}85^{t} = \dfrac{1}{2}.

Apply a logarithm to both sides:

t \cdot \log 0{,}85 = \log \tfrac{1}{2} \;\Longrightarrow\; t = \dfrac{\log(1/2)}{\log 0{,}85} \approx \dfrac{-0{,}3010}{-0{,}0706} \approx 4{,}27.

\boxed{\;t \approx 4{,}27\;\text{years (between 4 and 5 years).}\;}

The decreasing exponential $0{,}85^{t}$ is strictly monotonic, so the preimage is unique.

4 Reading from a graph

Suppose the graph of $f$ is the parabola $y = x^{2} - 4$. Answer using the graph as a visual reference (and check it algebraically).

a) What is the image of $-3$?

$f(-3) = (-3)^{2} - 4 = 9 - 4 = 5$.

\boxed{\;f(-3) = 5.\;}

b) How many preimages does the value $y = 5$ have? Compute them.

$x^{2} - 4 = 5 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3$. Two preimages.

\boxed{\;f^{-1}(5) = \{-3,\;3\}.\;}

c) How many preimages does the value $y = -4$ (vertex) have?

$x^{2} - 4 = -4 \Rightarrow x^{2} = 0 \Rightarrow x = 0$. A single preimage.

\boxed{\;f^{-1}(-4) = \{0\}.\;}

Geometrically, the line $y = -4$ is tangent to the parabola at the vertex.

d) How many preimages does the value $y = -5$ have?

$x^{2} - 4 = -5 \Rightarrow x^{2} = -1$, which has no real solutions.

\boxed{\;f^{-1}(-5) = \varnothing\;\;\text{(no preimage).}\;}

The line $y=-5$ is below the vertex of the parabola: it does not cut it.