Math Help Forum - View Single Post - I'm confused with some high school level algebra, help please?

masters · #8 July 1st, 2008, 05:39 PM

Here are some answers from your last page. Not much detail, but maybe it helps. Maybe some of the other guys here can finish up for you. Good luck.

1. If $f(x) = x^3 + 4$ , then compute $f^{-1}(5)$

$f^{-1}(x)$ means the inverse of $f(x)$

To find the inverse function, let the x and y change places. Then solve for y.

$f^{-1}(x)\rightarrow x=y^3+4$

$f^{-1}(x)\rightarrow y^3=x-4$

$f^{-1}(x)\rightarrow y=\sqrt[3]{x-4}$

Now find

$f^{-1}(5)=\sqrt[3]{5-4}=\sqrt[3]{1}=1$

2. Find the domain and range of $f(x) = \sqrt{x^2-25}$

$x^2-25\ge0$ over the Real number domain.

$D=\{x|x\leq -5 \ \ or \ \ x\geq 5\}$ or $D=(-\infty, -5] \cup [5, +\infty)$

$R=\{f(x)|f(x) \geq 0\}$

3. If $f(x) = 2x^2 + 4$ , find $f(x^4 – 1)$

$f(x^4-1)=2(x^4-1)^2+4\Longrightarrow2(x^8-2x^4+1)+4\Longrightarrow2x^8-4x^4+6$

4. If $z_1 = 3 + 2i$ and $z_2 = 4-5i$ , find

a. $(z_1)(z_2)$

$(3+2i)(4-5i)=12-7i+10=22-7i$

b. $\frac{z_1}{z_2}$

$\frac{3+2i}{4-5i}$

Multiply both numerator and denominator by the conjugate of $4-5i$

$\frac{3+2i}{4-5i}\cdot\frac{4+5i}{4+5i}=\frac{12+23i-10}{16+25}=\frac{2+23i}{41}$

5. Find $8^{-\frac{2}{3}}$

$8^{-\frac{2}{3}}=\frac{1}{8^{\frac{2}{3}}}=\frac{1}{\left(\sqrt[3]{8}^2\right)}=\frac{1}{4}$

The answer given for number 5 in your reference is wrong.