QuestionGiven the function $g(x)=1-x^2$ , find the domain, range, and evaluate $g(-1)$ and $g(2)$ . Describe the graph.

Studdy Solution

STEP 1

Assumptions1. The function is g(x)=1-x^ . We are asked to find the domain and range of $g$
3. We are asked to evaluate $g(-1)$ and $g()$ using the formula4. We are asked to evaluate $g(-1)$ and $g()$ using the graph of $g$

STEP 2

The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For the function $g(x)=1-x^2$ , we can substitute any real number for $x$ and get a real number for $g(x)$ . Therefore, the domain of $g$ is all real numbers.
$Domain\, of\, g = (-\infty, \infty)$

STEP 3

The range of a function is the set of all possible output values (y-values). For the function $g(x)=1-x^2$ , the highest value of $g(x)$ is achieved when $x=0$ , which gives $g(0)=1$ . As $x$ moves away from0 in either direction, $g(x)$ decreases, and there is no lower limit to how small $g(x)$ can be. Therefore, the range of $g$ is all real numbers less than or equal to1.
$Range\, of\, g = (-\infty,1]$

STEP 4

To evaluate $g(-1)$ , substitute $-1$ for $x$ in the formula.
$g(-1) =1 - (-1)^2$

STEP 5

Calculate $g(-1)$ .
$g(-1) =1 - (-1)^2 =1 -1 =0$

STEP 6

To evaluate $g(2)$ , substitute $2$ for $x$ in the formula.
$g(2) =1 -2^2$

STEP 7

Calculate $g(2)$ .
$g(2) =1 -2^2 =1 -4 = -3$

STEP 8

The graph of the function $g(x)=1-x^2$ is a downward-opening parabola with its vertex at the point (0,1). The x-intercepts (where $g(x)=0$ ) are at $x=-1$ and $x=1$ . The y-intercept (where $x=0$ ) is at $y=1$ . The graph extends infinitely to the left and right, and infinitely downward.

STEP 9

From the graph, we can see that $g(-)=$ and $g(2)=-3$ , which matches our calculations from the formula.

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