Answer to Question #208212 in Algebra for Zethu

Question

Suppose the functions f, g, h and l are as follows:

F(x)=4x^2-5x+1

g(x)=2√2-x/2-x

h(x)=-1/2x+3

l(x)= log4(x+3)-log4(x-2)

3.1 write down Df and solve the inequality f(x)<=0

3.2 Write down Dg, Dh and Dg+h.

3.3 Solve the equation g(x)=-4

3.4 Solve the equation 4^h(x)=8

3.5 Write down Dl and solve the equation l(x)= 1/2

Answer 1

$$f(x)=4x^2-5x+1$$

$$g(x)=\dfrac{2\sqrt{2-x}}{2-x}$$

$$h(x)=-\dfrac{1}{2x+3}$$

$$l(x)=\log_4(x+3)-\log_4(x-2)$$

3.1

$$f(x)=4x^2-5x+1$$

$Df:(-\infin, \infin)$

$$f(x)\leq 0=>4x^2-5x+1\leq 0$$

$$4x^2-4x-x+1\leq 0$$

$$4x(x-1)-(x-1)\leq 0$$

$$4(x-1)(x-\dfrac{1}{4})\leq 0$$

$$\dfrac{1}{4}\leq x\leq 1$$

$$x\in\big[\dfrac{1}{4}, 1\big]$$

3.2

$$g(x)=\dfrac{2\sqrt{2-x}}{2-x}$$

$$2-x\geq0\ and \ 2-x\not=0$$

Then $x<2$

$Dg: (-\infin, 2)$

$$h(x)=-\dfrac{1}{2x+3}$$

$$2x+3\not=0$$

Then $x\not=-\dfrac{3}{2}$

$Dh: (-\infin, -\dfrac{3}{2})\cup(-\dfrac{3}{2}, \infin)$

$$(g+h)(x)=\dfrac{2\sqrt{2-x}}{2-x}-\dfrac{1}{2x+3}$$

$D(g+h): (-\infin,- \dfrac{3}{2})\cup(-\dfrac{3}{2}, 2)$

3.3

$$g(x)=4$$

$$\dfrac{2\sqrt{2-x}}{2-x}=4, x<2$$

$$\sqrt{2-x}=\dfrac{1}{2}$$

$$2-x=(\dfrac{1}{2})^2$$

$$x=\dfrac{7}{4}$$

3.4

$$4^{h(x)}=8$$

$$4^{h(x)}=4^{3/2}$$

$$h(x)=\dfrac{3}{2}$$

$$-\dfrac{1}{2x+3}=\dfrac{3}{2}$$

$$6x+9=-2$$

$$6x=-11$$

$$x=-\dfrac{11}{6}$$

$x=-\dfrac{11}{6}$

3.5

$$l(x)=\log_4(x+3)-\log_4(x-2)$$

$$x+3>0\ and\ x-2 >0$$

$x>2$

$Dl: (2, \infin)$

$$l(x)=\dfrac{1}{2}$$

$$\log_4(x+3)-\log_4(x-2)=\dfrac{1}{2}, x>2$$

$$\log_4\dfrac{x+3}{x-2}=\dfrac{1}{2}$$

$$\dfrac{x+3}{x-2}=4^{1/2}$$

$$\dfrac{x+3}{x-2}=2$$

$$x+3=2x-4$$

$$x=7$$

$x=7$