Question

For the function y=-2+5sin(pi/12)(x-2)), what is the minimum value

Answer 1

Assuming the function is

$y=-2+5\sin\left(\dfrac\pi{12}(x-2)\right)$

recall that $-1\le\sin x\le1$, which means

$-1\le\sin\left(\dfrac\pi{12}(x-2)\right)\le1$
$\implies-5\le5\sin\left(\dfrac\pi{12}(x-2)\right)\le5$
$\implies-7\le-2+5\sin\left(\dfrac\pi{12}(x-2)\right)\le3$

and so the minimum value is -7.

Answer 2

Answer:

-7

Step-by-step explanation:

We are given that   a function

$y=-2+5 sin(\frac{\pi}{12}(x-2))$

We have to find the minimum value of y.

We know that range of sin x is [-1,1].

$-1\leq sin(\frac{\pi}{12}(x-2))\leq 1$

$-5\leq 5sin(\frac{\pi}{12}(x-2))\leq 5$

$-5-2\leq -2+5sin(\frac{\pi}{12}(x-2))\leq 5-2$

$-7\leq -2+5sin(\frac{\pi}{12}(x-2))\leq 3$

$-7\leq y\leq 3$

Maximum value of y=3

Minimum value of y=-7

Hence, the minimum value of given function =-7