EBUS PC12 Chapter 2 Function Operations and Coincs Exam-1

Step 1: Read the value of the function when \(x=3\). Step 2: From the graph, the point with \(x=3\) has \(y=-5\). Step 3: Therefore, \(f(3)=-5\). Answer: B. \(-5\)

Step 1: Use the vertical line test. Step 2: A relation is a function if every vertical line touches the graph at no more than one point. Step 3: Graph A passes the vertical line test, while the other graphs fail because at least one \(x\)-value has more than one \(y\)-value. Answer: Graph A

Step 1: The domain is the set of all possible \(x\)-values on the graph. Step 2: The graph starts at \(x=1\) with a closed dot, so \(1\) is included. Step 3: The graph ends at \(x=3\) with an open circle, so \(3\) is not included. Step 4: Therefore, the domain is \([1,3)\). Answer: C. \([1,3)\)

Step 1: The range is the set of all possible \(y\)-values on the graph. Step 2: From the graph, the lowest \(y\)-value is \(-5\). Step 3: The highest \(y\)-value is \(3\). Step 4: Both endpoint values are included, so the range is \(-5\le y\le 3\). Answer: C. \(-5\le y\le 3\)

Step 1: Compute I: \(g(3)=\frac{3-4}{2(3)}=-\frac{1}{6}\), so \((f\circ g)(3)=f\left(-\frac{1}{6}\right)=2\left(-\frac{1}{6}\right)^2=\frac{1}{18}\). Step 2: Compute II: \((f-g)(4)=f(4)-g(4)=2(4)^2-\frac{4-4}{2(4)}=32-0=32\). Step 3: Compute III: \(f(2)=2(2)^2=8\) and \(g(2)=\frac{2-4}{2(2)}=-\frac{1}{2}\), so \(\left(\frac{f}{g}\right)(2)=\frac{8}{-\frac{1}{2}}=-16\). Step 4: Compute IV: \((f+g)(1)=2(1)^2+\frac{1-4}{2(1)}=2-\frac{3}{2}=0.5\). Step 5: The lowest value is \(-16\). Answer: D. \(-16\)

Step 1: Expand \(g(x)=(2x-3)^2=4x^2-12x+9\). Step 2: Subtract: \(f(x)-g(x)=(x^2-6)-(4x^2-12x+9)\). Step 3: Combine like terms: \(x^2-6-4x^2+12x-9=-3x^2+12x-15\). Answer: B. \(-3x^2+12x-15\)

Step 1: Compute \(F(2x)=5-2x\). Step 2: Compute \(g(x-3)=3(x-3)+2=3x-9+2=3x-7\). Step 3: Set them equal: \(5-2x=3x-7\). Step 4: Solve: \(12=5x\), so \(x=\frac{12}{5}\). Answer: D. \(\frac{12}{5}\)

Step 1: Evaluate the inside function: \(g(1-\sqrt{3})=|(1-\sqrt{3})-3|=|-2-\sqrt{3}|\). Step 2: Since \(-2-\sqrt{3}<0\), \(|-2-\sqrt{3}|=2+\sqrt{3}\). Step 3: Now evaluate \(h(2+\sqrt{3})=(2+\sqrt{3})^2+2(2+\sqrt{3})\). Step 4: Expand: \((2+\sqrt{3})^2=7+4\sqrt{3}\) and \(2(2+\sqrt{3})=4+2\sqrt{3}\). Step 5: Add: \(7+4\sqrt{3}+4+2\sqrt{3}=11+6\sqrt{3}\). Answer: B. \(6\sqrt{3}+11\)

Step 1: First calculate \(g(4)=2-5(4)=2-20=-18\). Step 2: Then calculate \(g(g(4))=g(-18)=2-5(-18)\). Step 3: Simplify: \(2+90=92\). Answer: A. \(92\)

Step 1: For \(\left(\frac{f}{g}\right)(x)\), divide the \(y\)-value of \(f(x)\) by the \(y\)-value of \(g(x)\) at the same \(x\). Step 2: At \(x=3\), \(f(3)=10\) and \(g(3)=5\). Step 3: Therefore, \(\left(\frac{f}{g}\right)(3)=\frac{10}{5}=2\). Step 4: So the point is \((3,2)\). Answer: D. \((3,2)\)

Step 1: Composition \(g(f(x))\) means substitute \(f(x)\) into \(g(x)\). Step 2: Since \(g(x)=3x-5\), \(g(f(x))=3(f(x))-5\). Step 3: Substitute \(f(x)=x^2+4x\): \(g(f(x))=3(x^2+4x)-5\). Step 4: Expand: \(3x^2+12x-5\). Answer: A. \(3x^2+12x-5\)

Step 1: The composition is \(g(f(x))=g(\sqrt{x-2})=\sqrt{x-2}+3\). Step 2: The square root requires the radicand to be nonnegative. Step 3: Set \(x-2\ge 0\). Step 4: Solve to get \(x\ge 2\). Answer: A. \(x\ge 2\)

Step 1: Start with \(A(r)=4\pi r^2\). Step 2: Substitute \(r=\frac{C}{2\pi}\). Step 3: Then \(A(C)=4\pi\left(\frac{C}{2\pi}\right)^2\). Step 4: Simplify: \(A(C)=4\pi\cdot \frac{C^2}{4\pi^2}=\frac{C^2}{\pi}\). Answer: C. \(A(C)=\frac{C^2}{\pi}\)

Step 1: Switch \(x\) and \(y\): \(x=-2y^2+1\). Step 2: Solve for \(y\): \(x-1=-2y^2\). Step 3: Divide by \(-2\): \(y^2=\frac{x-1}{-2}\). Step 4: Take the square root: \(y=\pm\sqrt{\frac{x-1}{-2}}\). Answer: A. \(y=\pm\sqrt{\frac{x-1}{-2}}\)

Step 1: A point on \(f(x)\) has the form \((x,y)\). Step 2: A point on \(f^{-1}(x)\) swaps the coordinates to \((y,x)\). Step 3: Swap \((3,-1)\) to get \((-1,3)\). Answer: D. \((-1,3)\)

Step 1: Replacing \(y\) by \(-y\) changes the sign of the vertical coordinate. Step 2: Changing \(y\) to \(-y\) produces a reflection across the \(x\)-axis. Step 3: Therefore, the two graphs are reflected in the \(x\)-axis. Answer: D. They are reflected in the \(x\)-axis.

Step 1: The graph \(y=(x-4)^2+3\) is a parabola with vertex \((4,3)\). Step 2: A full parabola does not have an inverse that is a function because it fails the horizontal line test. Step 3: Restrict the domain to one side of the vertex. Step 4: From the choices, \(x\ge 4\) restricts the parabola to the right side, so the inverse is also a function. Answer: D. \(x\ge 4\)

Step 1: First find the composition: \((f\circ g)(x)=f(g(x))\). Step 2: Substitute \(g(x)=\frac{1}{3}x+1\) into \(f(x)=3x-2\): \(f(g(x))=3\left(\frac{1}{3}x+1\right)-2\). Step 3: Simplify: \(x+3-2=x+1\). Step 4: The inverse of \(y=x+1\) is found by switching variables: \(x=y+1\), so \(y=x-1\). Answer: B. \(x-1\)

Step 1: The inverse graph is the reflection of the original graph across the line \(y=x\). Step 2: The original graph is a downward-opening parabola, so its inverse is a sideways-opening relation. Step 3: The vertex of the original graph is approximately \((0,2)\), so the inverse vertex is approximately \((2,0)\). Step 4: Graph D matches this reflection. Answer: Graph D

Step 1: Find the inverse by switching \(x\) and \(y\): \(x=\frac{1}{y-4}\). Step 2: Solve for \(y\): \(x(y-4)=1\), so \(y-4=\frac{1}{x}\). Step 3: Therefore, the inverse is \(y=\frac{1}{x}+4\). Step 4: This graph has vertical asymptote \(x=0\) and horizontal asymptote \(y=4\), which matches Graph B. Answer: Graph B

Step 1: From the graph, the original \(x\)-intercepts are \(-7\), \(-2\), and \(5\). Step 2: For the transformed relation \(x=-f(y)\), use the transformation indicated in the question and compare intercept values. Step 3: The corresponding listed \(y\)-intercepts are the opposite signs of the original \(x\)-intercepts. Step 4: Thus the values are \(7\), \(2\), and \(-5\), which are listed as \(2,7,-5\). Answer: C. \(2,7,-5\)