Fundamentals of Functions (Leaving Cert Mathematics): Revision Notes

Worked Example: Function evaluation

Given: $f: x → x + 5$ and $g: x → x² - 1$

Find: (i) $f(3)$, (ii) $g(-3)$, (iii) $f(2k)$, (iv) $f(k + 1)$, (v) $g(3k)$, (vi) $g(k + 1)$

Solution:

• (i) $f(3) = 3 + 5 = 8$
• (ii) $g(-3) = (-3)² - 1 = 9 - 1 = 8$
• (iii) $f(2k) = 2k + 5$
• (iv) $f(k + 1) = (k + 1) + 5 = k + 6$
• (v) $g(3k) = (3k)² - 1 = 9k² - 1$
• (vi) $g(k + 1) = (k + 1)² - 1 = k² + 2k + 1 - 1 = k² + 2k$

Worked Example: Creating composite functions

Given: $f(x) = 2x + 5$ and $g(x) = x - 3$

To find $gf(x)$:

1. First apply $f$: $f(x) = 2x + 5$
2. Then apply $g$ to this result: $g(2x + 5) = (2x + 5) - 3 = 2x + 2$
3. Therefore: $gf(x) = 2x + 2$

To find $fg(x)$:

1. First apply $g$: $g(x) = x - 3$
2. Then apply $f$ to this result: $f(x - 3) = 2(x - 3) + 5 = 2x - 6 + 5 = 2x - 1$
3. Therefore: $fg(x) = 2x - 1$

Since $2x + 2 ≠ 2x - 1$, this confirms that $gf(x) ≠ fg(x)$.

Worked Example: Composite function calculations

Given: $f(x) = x + 3$ and $g(x) = x² - 1$

Find: (i) $fg(2)$, (ii) $gf(-1)$, (iii) $fg(x)$, (iv) $gf(x)$

Also find: the value of $x$ for which $fg(x) = gf(x)$

Solution:

(i) $fg(2)$:

• First find $g(2)$: $g(2) = 2² - 1 = 3$
• Then $f(3)$: $f(3) = 3 + 3 = 6$
• Therefore: $fg(2) = 6$

(ii) $gf(-1)$:

• First find $f(-1)$: $f(-1) = -1 + 3 = 2$
• Then $g(2)$: $g(2) = 2² - 1 = 3$
• Therefore: $gf(-1) = 3$

(iii) $fg(x)$:

• $fg(x) = f(x² - 1) = (x² - 1) + 3 = x² + 2$

(iv) $gf(x)$:

• $gf(x) = g(x + 3) = (x + 3)² - 1 = x² + 6x + 9 - 1 = x² + 6x + 8$

Finding where $fg(x) = gf(x)$:

• Set $x² + 2 = x² + 6x + 8$
• Subtract $x²$ from both sides: $2 = 6x + 8$
• Subtract 8: $-6 = 6x$
• Divide by 6: $x = -1$