Question

all guided lecture sheets (gls) must be submitted to gradescope in their entirety. that means uploading all pages of each gls. most problems will be graded for completion, so be sure to attempt every question—even if we didnt cover it in class. the last problem on each gls will be graded for accuracy. 6. given (f(x)=sqrt{2 - 4x}) and (g(x)=-\frac{3}{x}), find the following: a. ((gcirc f)(x)) b. the domain of ((gcirc f)(x)) in interval notation

Explanation:

Step1: Recall composition formula

$(g\circ f)(x)=g(f(x))$. Substitute $f(x)=\sqrt{2 - 4x}$ into $g(x)$.

Step2: Calculate $(g\circ f)(x)$

Since $g(x)=-\frac{3}{x}$, then $(g\circ f)(x)=-\frac{3}{\sqrt{2 - 4x}}$.

Step3: Find domain of $(g\circ f)(x)$

For the square - root in the denominator, we need $2-4x>0$ (because the denominator cannot be zero and the expression under the square - root must be non - negative). Solve $2-4x>0$. Subtract 2 from both sides: $-4x>-2$. Divide both sides by $- 4$ and reverse the inequality sign (since dividing by a negative number), we get $x <\frac{1}{2}$.

Answer:

a. $(g\circ f)(x)=-\frac{3}{\sqrt{2 - 4x}}$
b. $(-\infty,\frac{1}{2})$