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Functions & Curve Sketching | The Maths Tailor — UK Methods

Functions & Curve Sketching

10 methods · 25 questions

When a functional equation relates $f$ at two linked arguments (for example $x$ and $- x$ , or $x$ and some involution $K (x)$ with $K (K (x)) = x$ ), substitute that map into the equation to get a second equation in the same unknowns, then eliminate to solve for $f (x)$ explicitly.

Trigger: one equation, two appearances of $f$ at related arguments, and no other information.

Instances:
(i) replace $x$ by $- x$ to pair $f (x)$ with $f (- x)$ , then solve the $2 \times 2$ linear system.
(ii) apply an involution $K$ with $K (K (x)) = x$ (such as $K (x) = \frac{x + 1}{x - 1}$ ) so a second substitution closes the loop.
(iii) chain three substitutions when the map has order $3$ (such as $x \mapsto \frac{1}{1 - x}$ ), giving three equations in three values of $f$ .

Linked questions (1)

STEP 3 2016 — Q8

The function f satisfies, for all $x$ , the equation $f (x) + (1 - x) f (- x) = x^{2} .$ Show that $f (- x) + (1 + x) f (x) = x^{2}$ . Hence find $f (x)$ in terms of $x$ . You should verify that your function satisfies the original equation.
The function $K$ is defined, for $x \neq = 1$ , by $K (x) = \frac{x + 1}{x - 1} .$ Show that, for $x \neq = 1$ , $K (K (x)) = x$ . The function g satisfies the equation $g (x) + x g (\frac{x + 1}{x - 1}) = x (x \neq = 1) .$ Show that, for $x \neq = 1$ , $g (x) = \frac{2 x}{x ^{2} + 1}$ .
Find $h (x)$ , for $x \neq = 0$ , $x \neq = 1$ , given that $h (x) + h (\frac{1}{1 - x}) = 1 - x - \frac{1}{1 - x} (x \neq = 0, x \neq = 1) .$

Voir la correction

Part (i). We are given that

f (x) + (1 - x) f (- x) = x^{2}

for all

x

. Replace

x

- x

throughout:

f (- x) + (1 + x) f (x) = x^{2} .

This is the required identity. Now use it to eliminate

f (- x)

. From the original equation,

f (- x) = \frac{x ^{2} - f ( x )}{1 - x}

(for

x \neq = 1

). Substituting into the second equation:

\frac{x ^{2} - f ( x )}{1 - x} + (1 + x) f (x) = x^{2} .

Multiply through by

(1 - x)

x^{2} - f (x) + (1 - x^{2}) f (x) = x^{2} (1 - x) .

Collect the

f (x)

terms:

f (x) (- 1 + 1 - x^{2}) = x^{2} (1 - x) - x^{2}

, so

- x^{2} f (x) = - x^{3}

, giving

f (x) = x

. Verification:

f (x) + (1 - x) f (- x) = x + (1 - x) (- x) = x - x + x^{2} = x^{2}

. ✓

Part (ii). Showing $K (K (x)) = x$ : compute directly:

K (K (x)) = K (\frac{x + 1}{x - 1}) = \frac{\frac{x + 1}{x - 1} + 1}{\frac{x + 1}{x - 1} - 1} = \frac{( x + 1 ) + ( x - 1 )}{( x + 1 ) - ( x - 1 )} = \frac{2 x}{2} = x .

Finding $g (x)$ : the given equation is

g (x) + x g (\frac{x + 1}{x - 1}) = x

. Replace

x

K (x) = \frac{x + 1}{x - 1}

throughout:

g (\frac{x + 1}{x - 1}) + \frac{x + 1}{x - 1} g (x) = \frac{x + 1}{x - 1} .

Substitute the expression for

g (\frac{x + 1}{x - 1}) = \frac{x - g ( x )}{x}

(from the original equation, rearranged) into the new equation:

\frac{x - g ( x )}{x} + \frac{x + 1}{x - 1} g (x) = \frac{x + 1}{x - 1} .

Multiply through by

x (x - 1)

(x - 1) (x - g (x)) + x (x + 1) g (x) = x (x + 1) .

x (x - 1) - (x - 1) g (x) + x (x + 1) g (x) = x (x + 1) .

g (x) [x (x + 1) - (x - 1)] = x (x + 1) - x (x - 1) .

g (x) (x^{2} + 1) = x [(x + 1) - (x - 1)] = 2 x .

g (x) = \frac{2 x}{x ^{2} + 1} .

Part (iii). The key is to apply

x \mapsto \frac{1}{1 - x}

twice to generate two further equations. Let

ϕ (x) = \frac{1}{1 - x}

. Then

ϕ (ϕ (x)) = \frac{x - 1}{x}

and

ϕ (ϕ (ϕ (x))) = x

(the map has order 3).

The given equation is (call it E1):

h (x) + h (\frac{1}{1 - x}) = 1 - x - \frac{1}{1 - x}

.

Replace

x

\frac{1}{1 - x}

(E2):

h (\frac{1}{1 - x}) + h (\frac{x - 1}{x}) = \frac{x - 1}{x} - (1 - x)

.

Replace

x

\frac{x - 1}{x}

(E3):

h (\frac{x - 1}{x}) + h (x) = x - \frac{x - 1}{x}

.

Now compute E1

-

+

E3, in which the

h (\frac{1}{1 - x})

and

h (\frac{x - 1}{x})

terms cancel:

2 h (x) = (1 - x - \frac{1}{1 - x}) - (\frac{x - 1}{x} - 1 + x) + (x - \frac{x - 1}{x}) .

Simplifying the right-hand side gives

- 1

, so

h (x) = - \frac{1}{2} .