Restrict the domain to choose an inverse

f(x)=2/[1+(5x-1)^2]

Total marks: 10

Solution

(i)

Range

For all real $x$ , $f(x)=\dfrac{2}{1+(5x-1)^2}$ . Find its range.

Explore this idea

Hint

The denominator is smallest when $5x-1=0$ .
As $|x|\to\infty$ the fraction tends to 0 but never reaches it.

Answer · (i)

R_f=(0,2]

Explore the idea

A bell curve with maximum 2

f is symmetric about x = 1/5.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

F
Axis of symmetry

Ideas to try

Read off the range: which

y

-values does the curve take?

(ii)

One-one functions

Explain why $f$ has no inverse.

Explore this idea

Hint

Find two different inputs with the same output (horizontal-line test).

Answer · (ii)

f(0)=f(2/5)=1

Explore the idea

The horizontal line test

A function has an inverse only if every horizontal line meets its graph at most once.

Height of the horizontal linek = 1

Try an example

Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct intersections

Two distinct inputs give this output, so the function is not one-one.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Point	x	y
	-2.22045e-17	1
	0.4	1

F
Y = k
Intersections
Intersection selected in the coordinate table

Key idea

Two intersections at the same height mean two inputs share one output. The function is not one-one.

Ideas to try

For which

k

does the line meet the graph exactly once?

(iii)

Restricting a domain

The domain of $f$ is restricted to $x\ge k$ . Find the least $k$ for which $f^{-1}$ exists, and find $f^{-1}(x)$ .

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Hint

The restricted piece must be monotonic: keep one side of the peak.
Solve $y=f(x)$ for $x$ and choose the branch that matches the domain.
State the domain of the inverse.

Answer · (iii)

k=1/5

Explore the idea

Cut the domain until an inverse exists

Move the left boundary k. Keep only x ≥ k and watch when the two branches become one.

Original question: find the least k for a domain x ≥ k. Reset restores the starting experiment; the peak preset gives the boundary case.

Left boundary k0

Closed endpoint: the boundary is included unless the original domain excludes it.

Try an example

Not one-to-one: at least one reached output comes from two retained inputs.

Proposed input domain

[0, ∞)

Retained inputs

[0, ∞)

Reached outputs

(0, 2]

Original allowed inputs

(−∞, ∞)

These exclusions stay in force when a boundary changes. Arrows indicate continuation beyond the drawing window.

Both sides of the peak remain. A horizontal line can meet the retained graph twice.

The graph shows a finite window; arrows and interval notation describe the full set. Decimal endpoints are rounded.

(iv)

Graph of an inverse

Sketch the restricted $f$ and $f^{-1}$ on the same diagram, clearly showing their relationship.

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Hint

Use the same scale on both axes.
Draw $f$ from $\left(\frac15,2\right)$ downwards, and reflect it in $y=x$ .
Mark the end-points.

Restricted f and f⁻¹: equal coordinate unitsScroll across to see the full diagram.

Answer · (iv)

y=x

Explore the idea

Restricted f and its inverse

Drag the input. P = (x, f(x)) is on f and Q = (f(x), x) is on f⁻¹.

Point on f: input xx = 0.50

Reflection swaps the coordinates. The segment PQ is perpendicular to y = x, and its blue midpoint M lies on that line.

P = (x, f(x))

(0.5, 0.61538)

Q = (f(x), x)

(0.61538, 0.5)

on the inverse

Domain of f

x\ge\frac15

Range of f = domain of f⁻¹

0<y\le2

Midpoint M on y = x

(0.55769, 0.55769)

F restricted to x ≥ 1/5
F⁻¹
Y = x
PQ is perpendicular to y = x; its blue midpoint M lies on that line

Key idea

The inverse's vertical asymptote $x=0$ is the mirror of $f$ 's horizontal asymptote $y=0$ .

Ideas to try

Where do

f

and

f^{-1}

meet? Is it on

y=x

(v)

Fixed points

Show algebraically that a solution of $f(x)=f^{-1}(x)$ satisfies $25x^3-10x^2+2x-2=0$ .

Explore this idea

Hint

For a decreasing function the curves $y=f(x)$ and $y=f^{-1}(x)$ meet on $y=x$ .
So solve $f(x)=x$ .

Answer · (v)

f(x)=x\Rightarrow25x^3-10x^2+2x-2=0

Explore the idea

f meets y = x at one point

The curves f and f⁻¹ cross where f meets the line y = x.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

Restricted f
Y = x
Intersection

Ideas to try

Why is the cubic's only real root the right one?

Restrict the domain to choose an inverse

Solution

Range

Maximum

Lower end

A bell curve with maximum 2

One-one functions

A counterexample

Conclusion

The horizontal line test

Restricting a domain

Least $k$

Solve for $x$

Choose the branch

State the inverse

Cut the domain until an inverse exists

Proposed input domain

Retained inputs

Reached outputs

Original allowed inputs

Graph of an inverse

Restricted $f$

Reflect in $y=x$

Same scale

Restricted f and its inverse

Fixed points

Intersection on $y=x$

Substitute the rule

Clear the denominator

f meets y = x at one point

Maximum

Lower end

A counterexample

Conclusion

Least kkk

Solve for xxx

Choose the branch

State the inverse

Proposed input domain

Retained inputs

Reached outputs

Original allowed inputs

Restricted fff

Reflect in y=xy=xy=x

Same scale

Intersection on y=xy=xy=x

Substitute the rule

Clear the denominator

Least $k$

Solve for $x$

Restricted $f$

Reflect in $y=x$

Intersection on $y=x$