question 3(multiple choice worth 1 points)\n(secant, cosecant, and cotangent function behavior mc)\nfunction…

question 3(multiple choice worth 1 points)\n(secant, cosecant, and cotangent function behavior mc)\nfunction h is the reciprocal of function f. if function f is defined as f(t)=7 sec(11(t - π)) + 2, which describes the range of function h?\nthe range of function h is (-∞, -5∪9, ∞).\nthe range of function h is -7, 7.\nthe range of function h is -5, 9.\nthe range of function h is (-∞, -7∪7, ∞).

question 3(multiple choice worth 1 points)\n(secant, cosecant, and cotangent function behavior mc)\nfunction h is the reciprocal of function f. if function f is defined as f(t)=7 sec(11(t - π)) + 2, which describes the range of function h?\nthe range of function h is (-∞, -5∪9, ∞).\nthe range of function h is -7, 7.\nthe range of function h is -5, 9.\nthe range of function h is (-∞, -7∪7, ∞).

Answer

Answer:

The range of function $h$ is $(-\infty,-\frac{1}{5}]\cup[\frac{1}{9},\infty)$.

Explanation:

Step1: Recall range of secant function

The range of $y = \sec(x)$ is $(-\infty,- 1]\cup[1,\infty)$. For $y = 7\sec(11(t-\pi))$, when $\sec(11(t - \pi))$ takes its minimum value of $-1$, $y = 7\sec(11(t-\pi))=-7$; when $\sec(11(t - \pi))$ takes its maximum value of $1$, $y = 7\sec(11(t-\pi)) = 7$.

Step2: Find range of $f(t)$

Let $y = f(t)=7\sec(11(t - \pi))+2$. Then, when $7\sec(11(t - \pi))=-7$, $y=-7 + 2=-5$; when $7\sec(11(t - \pi))=7$, $y=7 + 2=9$. So the range of $f(t)$ is $(-\infty,-5]\cup[9,\infty)$.

Step3: Find range of $h(t)$

Since $h(t)=\frac{1}{f(t)}$, if $y\in(-\infty,-5]\cup[9,\infty)$, when $y\in(-\infty,-5]$, $\frac{1}{y}\in[-\frac{1}{5},0)$; when $y\in[9,\infty)$, $\frac{1}{y}\in(0,\frac{1}{9}]$. So the range of $h(t)$ is $(-\infty,-\frac{1}{5}]\cup[\frac{1}{9},\infty)$.