Question

use functions, $f(x)$, $g(x)$, $h(x)$ and $p(t)$ to answer the questions on the left.1. evaluate $f(-1)LXB0x = \underline{\quad\quad\quad}$3. evaluate $g(73)LXB1x = \underline{\quad\quad\quad}$5. evaluate $h(3)LXB2x = \underline{\quad\quad\quad}$7. evaluate $p(111)LXB3t = \underline{\quad\quad\quad}$attached are:- the graph of $f(x)$- $g(x): \\{(1,3.58), (20,74), (27,28), (29,50), (52,73), (61,-5), (67,27), (73,69), (74,85), (81,64)\\}$- $h(z) = -24z + 15$- the table of $p(t)$:| $t$ | 28 | 61 | 70 | 71 | 87 | 94 | 100 | 101 | 102 | 111 || --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- || $p(t)$ | 90 | 106 | 80 | 42 | 111 | 75 | 108 | 65 | 56 | 37 |

Explanation:

Step1: Evaluate $f(-1)$ from graph

From the graph of $f(x)$, when $x=-1$, $y=1$.

Step2: Find $x$ when $f(x)=-1$

From the graph of $f(x)$, when $y=-1$, $x=2$.

Step3: Evaluate $g(73)$ from set

Locate the pair in $g(x)$ where $x=73$: $(52,73)$ means $g(73)=52$.

Step4: Find $x$ when $g(x)=27$

Locate the pair in $g(x)$ where $y=27$: $(27,28)$ means $x=28$.

Step5: Evaluate $h(3)$ with formula

Substitute $x=3$ into $h(x)=-24x+15$:
$h(3)=-24(3)+15=-72+15=-57$

Step6: Find $x$ when $h(x)=303$

Set $-24x+15=303$, solve for $x$:
$-24x=303-15=288$
$x=\frac{288}{-24}=-12$

Step7: Evaluate $p(111)$ from table

Locate the row in $p(t)$ where $t=111$: $p(111)=111$.

Step8: Find $t$ when $p(t)=56$

Locate the value 56 in the $p(t)$ column: the corresponding $t=108$.

Answer:

$f(-1)=1$
$x=2$
$g(73)=52$
$x=28$
$h(3)=-57$
$x=-12$
$p(111)=111$
$t=108$