## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.3 - More on Functions and Their Graphs - Exercise Set - Page 197: 50

#### Answer

See the full explanation below.

#### Work Step by Step

(a) The input values to the function are known as the domain of the function. In the graph, usually x-values are the input values to the function, the graph of which is plotted. Hence, the domain of the function denotes the x-values of the graph. As the graph of $f$ extends to $-\infty$ in the negative x-direction and to $+\infty$ in the positive x-direction, the domain of f is given by $\left( -\infty ,\infty \right)$ Hence, the domain of $f$ is $\left( -\infty ,\infty \right)$. (b) The output values of the function are known as the range of the function. In the graph, usually y-values are the output values of the function, the graph of which was plotted. Hence, the range of the function denotes the y-values of the graph. As the graph of $f$ extends to $-\infty$ in the negative y-direction and to $4$ in the positive y-direction, the range of f is $\left( -\infty ,4 \right]$. Hence, the range of $f$ is given by $\left( -\infty ,4 \right]$. (c) The x-intercept is defined as the point where the graph cuts the x-axis. As the graph of $f$ cuts the x-axis at−4 and 4, the x-intercepts are $-\text{4 and 4}$ Hence, the x-intercepts of $f$ are $-\text{4 and 4}$. (d) The y-intercept is defined as the point where the graph cuts the y-axis. As the graph of $f$ cuts they-axis at 1 only, the y-intercept is $1$ Hence, the y-intercept of $f$ is $1$. (e) The function is said to be increasing when with the increase in x-values the corresponding y-values of the graph of the function also increases. The value of the provided function increases in the intervals $\left( -\infty ,-2 \right)\text{ and }\left( 0,3 \right)$. Hence, the intervals in which $f$ is increasing are $\left( -\infty ,-2 \right)\text{ and }\left( 0,3 \right)$. (f) The function is said to be decreasing when the y-values of the graph of the function decrease with the increase in x-values. The value of provided function decreases in the intervals $\left( -2,0 \right)\text{ and }\left( 3,\infty \right)$. Hence, the intervals in which $f$ is decreasing are $\left( -2,0 \right)\text{ and }\left( 3,\infty \right)$. (g) The given function has values less than or equal to zero, that is, the value of the provided function $f\left( x \right)\le 0$ is for the values of x which lies in the intervals $\left( -\infty \text{,}-\text{4} \right]\text{ and }\left[ \text{4,}\infty \right)$. Hence, the values of x for which $f\left( x \right)\le 0$ lies in the intervals $\left( -\infty \text{,}-\text{4} \right]\text{ and }\left[ \text{4,}\infty \right)$. (h) The relative minimum of a function is defined as that value of x at which the y-value will be minimum as compared to all other points. Thus, the value of the provided function has maximum value at $x=-2$. Hence, the number at which $f$ has a relative maximum is $-2$. (i) The relative minimum of a function is defined as that value of x at which the y-value will be minimum as compared to all other points. The value of provided function has maximum value at $x=-2$ and the maximum value of $f$ at $x=-2$ is $4$. Hence, the relative maximum of $f$ is $4$. (j) As the value of yis 4 when x is −2, the required value is $f\left( -2 \right)=4$ Hence, the value of $f\left( -2 \right)$ is $4$. (k) There are two points on the x-axis for which the y-value or the f(x) is $0$. So, the corresponding values of x are $f\left( x \right)=0 \text{ for } x=-4 \text{ and } x=4$ Hence, the values of x for which $f\left( x \right)=0$ are $-4 \text{ and } 4$. (l) An even function is the function having symmetry with the y-axis An odd function is the function having symmetry with the origin. As the provided graph has symmetry with none, so the provided function is neither even nor odd. Hence, the provided function $f$ is neither even nor odd.

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