Question

1. (f∘g)(1)=34. g(g(−1))=35. (g∘f)(1)=

Explanation:

Step1: Find \(g(1)\) and \(f(g(1))\)

From the graph, assume \(g(x)\) is a linear - like function (by observing the graph's shape for \(g(x)\)). Let's first find \(g(1)\).
If we assume \(g(x)\) passes through \((0,1)\) and \((2,0)\), the slope \(m=\frac{0 - 1}{2-0}=-\frac{1}{2}\), and the equation \(g(x)=-\frac{1}{2}x + 1\). Then \(g(1)=-\frac{1}{2}(1)+1=\frac{1}{2}\).
Now, for \(f(x)\), assume \(f(x)=-x^{2}+2x + 1\) (by using the vertex form \(y=a(x - h)^{2}+k\), vertex \((h,k)=(1,2)\), \(a=-1\)). Then \(f(g(1))=f(\frac{1}{2})\).
Substitute \(x = \frac{1}{2}\) into \(f(x)\): \(f(\frac{1}{2})=-(\frac{1}{2})^{2}+2(\frac{1}{2})+1=-\frac{1}{4}+1 + 1=\frac{7}{4}\).

Step2: Find \(g(-1)\) and \(g(g(-1))\)

Substitute \(x=-1\) into \(g(x)=-\frac{1}{2}x + 1\), \(g(-1)=-\frac{1}{2}(-1)+1=\frac{1 + 2}{2}=\frac{3}{2}\).
Then \(g(g(-1))=g(\frac{3}{2})\). Substitute \(x = \frac{3}{2}\) into \(g(x)\): \(g(\frac{3}{2})=-\frac{1}{2}(\frac{3}{2})+1=-\frac{3}{4}+1=\frac{1}{4}\).

Step3: Find \(f(1)\) and \(g(f(1))\)

Substitute \(x = 1\) into \(f(x)=-x^{2}+2x + 1\), \(f(1)=-1^{2}+2(1)+1=2\).
Then \(g(f(1))=g(2)\). Substitute \(x = 2\) into \(g(x)=-\frac{1}{2}x + 1\), \(g(2)=-\frac{1}{2}(2)+1=0\).

Answer:

1. \(\frac{7}{4}\)
2. \(\frac{1}{4}\)
3. \(0\)