Answer:
a) The domain of the function is [tex]x \geq 0\,s[/tex] [tex]\wedge[/tex] [tex]x \leq 5.112\,s[/tex]. [tex][0\,s, 5.112\,s][/tex], [tex]\forall x \in \mathbb{R}[/tex], b) The range of the function is [tex]0\,m \leq y \leq 100\,m[/tex]. [tex][0\,m,100\,m][/tex], [tex]\forall y\in \mathbb{R}[/tex], c) The ball is 73 meters off of the ground at x = 3 seconds.
Step-by-step explanation:
The complete statement is: A ball is thrown upward off of a 100 meter cliff with an initial velocity of 6 m/s. The function [tex]f(x) = -5\cdot x^{2} + 6\cdot x + 100[/tex] represents this situation where x is time and y is the distance off of the ground.
a) What domain does the function make sense?
b) What range does the function make sense ?
c) How far off the ground is the ball at time x = 3 seconds?
a) Let [tex]x[/tex] and [tex]f(x)[/tex] be the time, measured in seconds, and the distance of the ground, measured in meters, respectively. Time is a positive variable, so domain corresponds to the interval when [tex]f(x) \geq 0[/tex] and [tex]t \geq 0[/tex]. That is:
[tex]-5\cdot x^{2} + 6\cdot x + 100 \geq 0[/tex]
[tex]-(x-5.112\,s)\cdot (x+3.912\,s) \geq 0[/tex]
Therefore, the domain of the function is [tex]x \geq 0\,s[/tex] [tex]\wedge[/tex] [tex]x \leq 5.112\,s[/tex]. [tex][0\,s, 5.112\,s][/tex], [tex]\forall x \in \mathbb{R}[/tex]
b) The distance off of the ground is also a positive variable, where ball is thrown upward at a height of 100 meters and hits the ground at a height of 0 meters. Hence, the range of the function is [tex]0\,m \leq y \leq 100\,m[/tex]. [tex][0\,m,100\,m][/tex], [tex]\forall y\in \mathbb{R}[/tex]
c) The distance of the ball off of the ground at x = 3 seconds is found by evaluating the function:
[tex]f(3\,s) = -5\cdot (3\,s)^{2} + 6\cdot (3\,s) + 100[/tex]
[tex]f(3\,s) = 73\,m[/tex]
The ball is 73 meters off of the ground at x = 3 seconds.
