Answer to Question #194247 in Calculus for Maureen

Question

Let f(x)=x/3xsquared+1 and g(x)=square root of 1-x

Find

1. f+g

2. F-g

3. F/g

4. F of g

5. G of f

6. State the domain of g of f

Answer 1

$Given, f(x)=\frac{x}{3x^2+1} and g(x)=\sqrt{1-x}.\newline 1.\newline f+g=\frac{x}{3x^2+1} +\sqrt{1-x}\newline =\frac{x+\sqrt{1-x}}{3x^2+1} \newline 2.\newline f-g=\frac{x}{3x^2+1} -\sqrt{1-x}\newline =\frac{x-\sqrt{1-x}}{3x^2+1} \newline 3.\newline \frac{f}{g}=\frac{\frac{x}{3x^2+1} }{\sqrt{1-x}}\newline =\frac{x}{(3x^2+1)\sqrt{1-x}} \newline 4.\newline f\omicron g=f(g(x))=f(\sqrt{1-x})=\frac{\sqrt{1-x}}{3(\sqrt{1-x})^2+1} =\frac{\sqrt{1-x}}{3(1-x)^2+1} =\frac{\sqrt{1-x}}{3(1-2x+x^2)+1} =\frac{\sqrt{1-x}}{3x^2-6x+4} \newline 5.\newline g\omicron f=g(f(x))=g(\frac{x}{3x^2+1})=\sqrt{1-\frac{x}{3x^2+1}}=\sqrt{\frac{3x^2+1-x}{3x^2+1}} =\sqrt{\frac{3x^2-x+1}{3x^2+1}} \newline 6.\newline \text{Domain of square root function is Positive real numbers.}\newline \text{Therefore, denominator>0.}\newline 3x^2+1<0\newline \text{Since, }3x^2+1 \text{is always positive.} \newline \text{Therefore,domain of } g\omicron f=\sqrt{\frac{3x^2-x+1}{3x^2+1}}\text{ is real number} i.e., x\in \mathbb{R}. \newline$