Question

1. without using your calculator, evaluate each of the following given the function definitions and input values:

(a) (f(x)=3x + 7)
(f(-4)=3(-4)+7)
(f(2)=3(2)+7 = 13)
(b) (g(x)=3x^{2})
(g(2)=3(2)^{2}=12)
(g(-3)=3(-3)^{2}=27)
(c) (h(x)=sqrt{x - 5})
(h(41)=sqrt{41 - 5}=6)
(h(14)=sqrt{14 - 5}=3)

1. using store on your calculator, evaluate each of the following more complex functions:

(a) (f(x)=\frac{3x^{2}-5}{4x + 10})
(f(-5)=\frac{3(-5)^{2}-5}{4(-5)+10}=-7)
(f(0)=\frac{3(0)^{2}-5}{4(0)+10}=-\frac{1}{2})
(b) (g(x)=\frac{sqrt{25 - x^{2}}}{x})
(g(4)=\frac{sqrt{25-(4)^{2}}}{4}=\frac{3}{4})
(g(-3)=\frac{sqrt{25-(-3)^{2}}}{-3}=-\frac{4}{3})
(c) (h(x)=30(1.2)^{x})
(h(3)=30(1.2)^{3}=51.84)
(h(0)=30(1.2)^{0}=30)

1. the function (y = g(x)) is shown graphed below. answer the following questions based on the graph.

(a) give values for (g(-2),g(0),g(3)) and (g(7)).
(b) for what values of (x) will (g(x)=0)? circle the points on the graph that help answer this question.
(c) find the value of (g(9)-g(-1)).
(d) what does the calculation in (c) tell you about the function on the interval (-1leq xleq9)?

Explanation:

Step1: Evaluate $f(x) = 3x + 7$ at $x=-4$ and $x = 2$

Substitute $x=-4$ into $f(x)$: $f(-4)=3(-4)+7=-12 + 7=-5$. Substitute $x = 2$ into $f(x)$: $f(2)=3(2)+7=6 + 7 = 13$.

Step2: Evaluate $g(x)=3x^{2}$ at $x = 2$ and $x=-3$

Substitute $x = 2$ into $g(x)$: $g(2)=3(2)^{2}=3\times4 = 12$. Substitute $x=-3$ into $g(x)$: $g(-3)=3(-3)^{2}=3\times9=27$.

Step3: Evaluate $h(x)=\sqrt{x - 5}$ at $x = 41$ and $x = 14$

Substitute $x = 41$ into $h(x)$: $h(41)=\sqrt{41-5}=\sqrt{36}=6$. Substitute $x = 14$ into $h(x)$: $h(14)=\sqrt{14 - 5}=\sqrt{9}=3$.

Step4: Evaluate $f(x)=\frac{3x^{2}-5}{4x + 10}$ at $x=-5$ and $x = 0$

Substitute $x=-5$ into $f(x)$: $f(-5)=\frac{3(-5)^{2}-5}{4(-5)+10}=\frac{3\times25-5}{-20 + 10}=\frac{75 - 5}{-10}=\frac{70}{-10}=-7$. Substitute $x = 0$ into $f(x)$: $f(0)=\frac{3(0)^{2}-5}{4(0)+10}=\frac{-5}{10}=-\frac{1}{2}$.

Step5: Evaluate $g(x)=\frac{\sqrt{25 - x^{2}}}{x}$ at $x = 4$ and $x=-3$

Substitute $x = 4$ into $g(x)$: $g(4)=\frac{\sqrt{25-4^{2}}}{4}=\frac{\sqrt{25 - 16}}{4}=\frac{\sqrt{9}}{4}=\frac{3}{4}$. Substitute $x=-3$ into $g(x)$: $g(-3)=\frac{\sqrt{25-(-3)^{2}}}{-3}=\frac{\sqrt{25 - 9}}{-3}=\frac{\sqrt{16}}{-3}=-\frac{4}{3}$.

Step6: Evaluate $h(x)=30(1.2)^{x}$ at $x = 3$ and $x = 0$

Substitute $x = 3$ into $h(x)$: $h(3)=30(1.2)^{3}=30\times1.728 = 51.84$. Substitute $x = 0$ into $h(x)$: $h(0)=30(1.2)^{0}=30\times1=30$.

Step7: Analyze the graph of $y = g(x)$ for $g(-2),g(0),g(3),g(7)$

From the graph, $g(-2)=3$, $g(0)=0$, $g(3)=3$, $g(7)=0$.

Step8: Find $x$ - values for $g(x)=0$ from the graph

From the graph, $g(x)=0$ when $x=-3,1,5,7$.

Step9: Calculate $g(9)-g(-1)$

From the graph, $g(9)=8$ and $g(-1)=4$, so $g(9)-g(-1)=8 - 4=4$.

Step10: Interpret the result of $g(9)-g(-1)$

The function $y = g(x)$ has a net increase of 4 over the interval $-1\leq x\leq9$.

Answer:

1.

• (a) $f(-4)=-5,f(2)=13$
• (b) $g(2)=12,g(-3)=27$
• (c) $h(41)=6,h(14)=3$

2.

• (a) $f(-5)=-7,f(0)=-\frac{1}{2}$
• (b) $g(4)=\frac{3}{4},g(-3)=-\frac{4}{3}$
• (c) $h(3)=51.84,h(0)=30$

3.

• (a) $g(-2)=3,g(0)=0,g(3)=3,g(7)=0$
• (b) $x=-3,1,5,7$
• (c) $4$
• (d) The function $y = g(x)$ has a net increase of 4 over the interval $-1\leq x\leq9$.