Determine if the following functions are valid, and, if they are, whether they are one-to-one or many-to-one: - $\mathrm{f}(x) = \pm\sqrt{x}$
- $\mathrm{f}(x) = \dfrac{1}{x}$
- one to many, not a function
- one to one, function
Find the range of the following functions: - $\mathrm{f}(x) = x^2-5, \quad x \in \{-2,0,2\}$
- $\mathrm{f}(x) = \dfrac{2}{1-x}, \quad x \in \{-2,0,2\}$
- $\{-5,-1\}$
- $\{-2,\frac{2}{3},2\}$
A function is linear from $(-10,14)$ to $(-4,2)$, and from $(-4,2)$ to $(6,27)$. Solve $\mathrm{f}(x) = 12$
$x = -9, 0$
$$\mathrm{f}(x) = 2x^2 + 8x - 1$$ Find the largest possible domain for $\mathrm{f}$ such that it has an inverse.
$x \geqslant -2$ or $x \leqslant -2$
$$\mathrm{f}:x\mapsto x^2+2x-2$$ Find $\mathrm{f}:\{1, 2\}$
$\{1, 6\}$
Sketch the following graphs: - $y=|x-1|$
- $y=|4x-3|$
Sketch the following graphs: - $y=|7-x|$
- $y=-|x-1|$
Given $\mathrm{f}(x) = \dfrac{1}{x-2}$ and $\mathrm{g}(x) = 3x+4$, solve $\mathrm{gf}(x)=16$
$\dfrac{9}{4}$
Given $\mathrm{f}(x) = 5-2x$, solve $$\mathrm{f}^2(x) - (\mathrm{f}(x))^2 = 0$$
$\dfrac{6\pm\sqrt{6}}{2}$
$$\mathrm{f}(x) = \dfrac{x+6}{x+2}$$ $$\mathrm{g}(x) = 7-2x^2$$ - Find the range of $\mathrm{g}$.
- Solve $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$.
- $\mathrm{g}(x) \leqslant 7$
- $x = -3, 2$
The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = x^2+4x+9, \quad x\geqslant a$ - State the smallest value of $a$ for which $\mathrm{f}^{-1}(x)$ exists.
- Find $\mathrm{f}^{-1}(x)$ and state its domain.
- $-2$
- $\mathrm{f}^{-1}(x) = \sqrt{x-5}-2, \quad x \geqslant 5$
Solve: - $|3x+1|=5$
- $\left|\dfrac{x-5}{2}\right|=1$
Solve: - $\left|\dfrac{x}{6}-1\right|=3$
- $|3x-5|=11-x$
Solve: - $3+\mathrm{e}^{2x}=7\mathrm{e}^{2x}$
- $\mathrm{e}^{2x}=\mathrm{e}^x+2$
- $-\frac{1}{2}\ln2$
- $\ln 2$
Solve: - $|x-9| < 2$
- $\left|3-\dfrac{3x}{4}\right| > \dfrac{1}{4}$
- $7 < x < 11$
- $x < \dfrac{11}{3}$ or $x > \dfrac{13}{3}$
Solve: - $|2x+9| > 14 - x$
- $-|3x+4| \leqslant 2x-9$
- $x > \dfrac{5}{3}$ or $x < -23$
- $x \geqslant 1$ or $x \leqslant -13$
Solve: - $3|4x-3|-1=14$
- $3-|2x+3|=1$
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- $-\frac{1}{2}, 2$
- $-\frac{5}{2}, -\frac{1}{2}$
$$\mathrm{f}(x) = 2x^2-1, \quad x \geqslant a$$ - Given that $\mathrm{f}^{-1}$ exists, suggest a suitable domain for $\mathrm{f}$.
- Find a solution to $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$.
The function $\mathrm{f}$ has domain $-5 \leqslant x \leqslant 14$ and is linear from $(-5,-8)$ to $(0,12)$ and from $(0,12)$ to $(14,5)$. - Find $\mathrm{ff}(0)$.
- Given $\mathrm{g}:x\mapsto\dfrac{2x-5}{10-x}$, find $\mathrm{fg}(7)$.
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- $6$
- $\frac{21}{2}$
Solve: - $\ln(3x-2) - \ln 2 = \ln(x+4)$
- $2\ln(x) = \ln(2x+3)$
Given $$\mathrm{f}:x\mapsto x^2+3$$ $$\mathrm{g}:x\mapsto 2x+2$$ solve $$\mathrm{fg}(x) = 2\mathrm{gf}(x)+15$$
$3$
Solve - $3|x| + 5 = 5|x|-1$
- $|4-x| = |4x-1|$
Solve: - $|2x-3| > |x+3|$
- $|4x-1| < |x+1|$
- $x < 0$ or $x > 6$
- $0 < x < \dfrac{2}{3}$
Solve: - $6|x| \geqslant |2-3x|$
- $|x-3| > 2|x+1|$
- $x \geqslant \dfrac{2}{9}$ or $x \leqslant -\dfrac{2}{3}$
- $-5 < x < \dfrac{1}{3}$