Answer to Question #158147 in Algebra for Geek

Question

1. Let f(x) = √ x + 1 − 1/4−x² Compute the following

i. f(0)

ii. f(-3)

iii. f(2)

iv. f(-1)

v. f(3)

vi. Domain(f(x)) 

2. (a) Let f( x/x−2 ) = 3x + 4 find f(x).

(b) Compute difference quotient of the function g(x) = √ x 2 − 9, and simplify your answer.

3. Let f(x) = 1 − x, g(x) = x 2 + bx + c. find b and c such that fog(x) = −X²+ 5x + 4.

Answer 1

$\mathrm{Given : f(x)=\sqrt{x+1}-\dfrac{1}{4-x^2}}\\ \;\\ \textup{i. f(0) = }\sqrt{0+1}-\dfrac{1}{4-0^2}=1-\dfrac{1}{4}=\dfrac{3}{4}\\ \; \\ \textup{ii. f(-3) = }\sqrt{-3+1}-\dfrac{1}{4-(-3)^2}=\sqrt{-2}+\dfrac{1}{5}=\textup{does not exist}\\ \; \\ \textup{iii. f(2) = }\sqrt{2+1}-\dfrac{1}{4-2^2}=\sqrt{3}-\dfrac{1}{0};\textup{does not exist}\\ \; \\$

$\textup{iv. f(-1) = }\sqrt{-1+1}-\dfrac{1}{4-(-1)^2}=0-\dfrac{1}{4-1}=-\dfrac{1}{3}\\ \; \\ \textup{v. f(3) = }\sqrt{3+1}-\dfrac{1}{4-(3)^2}=\sqrt{4}+\dfrac{1}{5}=2+\dfrac{1}{5}=\dfrac{11}{5}\\ \; \\$

$\textup{vi. Domain(f(x)) :-}\\ \; \\ \textup{Whenever we have a function with a square root, we know that the square root }\\ \textup{has to be bigger than or equal to 0.}\\ \; \\ \sqrt{x+1}\geq 0 \\ \; \\ x+1\geq 0 \\ \; \\ x\geq -1 \\ \; \\ \textup{For the second part, the denominator must be non-zero. So,}\\ \; \\ 4-x^2\neq 0 \\ \; \\ x^2\neq 4\\ \; \\ x\neq 2, x\neq -2\\ \; \\ \textup{So our domain is }[-1,\infty)-\{2\}\\ \; \\$

$2.\;\textup{(a) Given : f}\left(\dfrac{x}{x-2} \right)=3x+4 \\ \; \\ \implies \textup{Let }t=\left(\dfrac{x}{x-2} \right)\\ \; \\ \implies t=\left(\dfrac{x-2+2}{x-2} \right)=1+\dfrac{2}{x-2}\\ \; \\ \implies t-1=\dfrac{2}{x-2}\\ \; \\ \implies x-2=\dfrac{2}{t-1}\\ \; \\ \implies x=2+\dfrac{2}{t-1}\\ \; \\ \therefore f(t)=3\left(2+\dfrac{2}{t-1} \right)+4\\ \ \\ \implies f(t)=6+\dfrac{6}{t-1}+4=\dfrac{6}{t-1}+10\\ \; \\ \textup{Changing the variable from t to x,}\\ \; \\ \therefore f(x)=\dfrac{6}{x-1} +10\\ \ \\$

$\textup{(b) Given : }g(x)=\sqrt{x^2-9}\\ \; \\ \textup{To find : difference quotient }\;\dfrac{g(x+h)-g(x)}{h}\\ \;\\ \textup{Here, }g(x+h)=\sqrt{(x+h)^2-9}\\ \; \\ \therefore \dfrac{g(x+h)-g(x)}{h}=\dfrac{\sqrt{(x+h)^2-9}-\sqrt{x^2-9}}{h}\\ \; \\ \textup{Rationalizing the numerator and denominator, we get}\\\;\\ \therefore \dfrac{g(x+h)-g(x)}{h}=\dfrac{\sqrt{(x+h)^2-9}-\sqrt{x^2-9}}{h}\times \dfrac{\sqrt{(x+h)^2-9}+\sqrt{x^2-9}}{\sqrt{(x+h)^2-9}+\sqrt{x^2-9}}\\ \; \\ \textup{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\dfrac{(x+h)^2-9-(x^2-9)}{h(\sqrt{(x+h)^2-9}+\sqrt{x^2-9})}\\ \; \\ \textup{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\dfrac{(x+h)^2-x^2}{h(\sqrt{(x+h)^2-9}+\sqrt{x^2-9})}\\ \; \\$

$\textup{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\dfrac{(x+h-x)(x+h+x)}{h(\sqrt{(x+h)^2-9}+\sqrt{x^2-9})}\\ \; \\ \textup{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\dfrac{h(2x+h)}{h(\sqrt{(x+h)^2-9}+\sqrt{x^2-9})}\\ \; \\ \textup{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\dfrac{2x+h}{\sqrt{(x+h)^2-9}+\sqrt{x^2-9}}\\ \; \\$

$3. \textup{ Given : }f(x) = 1 − x, g(x) = x^2 + bx + c. \\ \; \\ \textup{To find : b and c such that }fog(x) = −x^2+ 5x + 4.\\ \; \\ \textup{Here, }fog(x)=f(g(x))=f(x^2+bx+c)=1-(x^2+bx+c)=-x^2-bx+(1-c)\\ \; \\ \because fog(x)=−x^2+ 5x + 4 \\ \; \\ \implies -x^2-bx+(1-c)=-x^2+5x+4\\ \; \\ \textup{Comparing both sides , we get}\\ \; \\ -b=5,\;1-c=4\\ \; \\ \therefore b=-5,\;\;c=-3$