A function f is defined by f(x) 5-2sin2x For 0<x<pie a)find the .range of f. b)sketch the graph of y=f (x)?

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Answer 1 · 2017-11-18T06:48:43

The range is $f (x) \in [3, 7]$ $f(x) in [3,7]$ . See the graph below

To find the range, proceed as follows :

$- 1 \leq sin x \leq 1$ $-1 <= sinx <=1$

$- 1 \leq sin 2 x \leq 1$ $-1 <= sin2x <=1$

$1 \geq - sin 2 x \geq - 1$ $1 >= -sin2x >=-1$

$2 \geq - 2 sin 2 x \geq - 2$ $2 >= -2sin2x >=-2$

$2 + 5 \geq (5 - 2 sin 2 x) \geq - 2 + 5$ $2+5 >= (5-2sin2x )>=-2+5$

$3 \leq f (x) = (5 - 2 sin 2 x) \leq 7$ $3<= f(x)=(5-2sin2x) <=7$

Therefore,

the range is $f (x) \in [3, 7]$ $f(x) in [3,7]$

To sketch the graph in the domain $x \in (0, π)$ $x in (0, pi)$

Calculate the following values

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$

$a a a a$ $color(white)(aaaa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $5$ $5$

$a a a a$ $color(white)(aaaa)$ $\frac{π}{4}$ $pi/4$ $a a a a$ $color(white)(aaaa)$ $3$ $3$

$a a a a$ $color(white)(aaaa)$ $\frac{π}{2}$ $pi/2$ $a a a a$ $color(white)(aaaa)$ $5$ $5$

$a a a a$ $color(white)(aaaa)$ $\frac{3}{4} π$ $3/4pi$ $a a a a$ $color(white)(aaaa)$ $7$ $7$

$a a a a$ $color(white)(aaaa)$ $π$ $pi$ $a a a a$ $color(white)(aaaa)$ $5$ $5$

$5 - 2 sin 2 x = 6$ $5-2sin2x=6$

$2 sin 2 x = - 1$ $2sin2x=-1$

$sin 2 x = - \frac{1}{2}$ $sin2x=-1/2$

$2 x = 7 \frac{π}{6}$ $2x=7pi/6$ , $\Rightarrow$ $=>$ , $x = \frac{7}{12} π$ $x=7/12pi$

$2 x = \frac{11}{6} π$ $2x=11/6pi$ , $\Rightarrow$ $=>$ , $x = \frac{11}{12} π$ $x=11/12pi$

See the graph below

graph{(y-5+2sin(2x))(y-6)=0 [-10, 10, -5, 5]}