1.Let f(x)={x+a,x≤0x+4,x>0f(x) = \begin{cases} \sqrt{x+a}, & x \le 0 \\ x+4, & x > 0 \end{cases}f(x)={x+a,x+4,x≤0x>0 and g(x)={x+1,x<0(x−4)2+b,x≥0g(x) = \begin{cases} x+1, & x < 0 \\ (x-4)^2+b, & x \ge 0 \end{cases}g(x)={x+1,(x−4)2+b,x<0x≥0 are continuous on R\mathbb{R}R. Then (g∘f)(2)+(f∘g)(−2)(g \circ f)(2) + (f \circ g)(-2)(g∘f)(2)+(f∘g)(−2) is equal to:Login to continueOnly logged in users canattempt or see the solution.
