Question

range:
d) f(x)=2sinx + 1
domain:
range:

Explanation:

Step1: Recall range of sin(x)

The range of the basic sine - function $y = \sin x$ is $[- 1,1]$, i.e., $-1\leqslant\sin x\leqslant1$.

Step2: Manipulate the inequality for $2\sin x+1$

Multiply the inequality for $\sin x$ by 2: $-2\leqslant2\sin x\leqslant2$. Then add 1 to each part of the inequality: $-2 + 1\leqslant2\sin x+1\leqslant2 + 1$.

Step3: Simplify the inequality

We get $-1\leqslant2\sin x + 1\leqslant3$.

Answer:

$[-1,3]$