Question

Find the maximum and minimum values of |3sinx+2| without using derivatives

Answer 1

Max  of$\mid 3sinx+2\mid$=5

Min of $\mid 3sinx+2\mid$=1

Step-by-step explanation:

Given:

$f(x)=\mid{3sinx+2}\mid$

We know that

$-1\leq sinx\leq 1$

$-3\leq 3sinx\leq 3$

$-3+2\leq 3sinx +2\leq 3+2$

$-1\leq 3sinx +2\leq 5$

Maximum value of f(x)=5

We know that

$\mid 3sinx+2\mid\geq 0$

Substitute x=$-\frac{\pi}{2}$

$\mid{3sin(-\frac{\pi}{2}+2}\mid$

$1$

Because

$sin(-\frac{\pi}{2})=-1$

Hence, the maximum value of f(x)=5 and minimum value of f(x)=1