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

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

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

$x>2$

$Dl: (2, \infin)$

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

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

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

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

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

$x+3=2x-4$

$x=7$

$x=7$