What is the maximum value of y = 5 - 4 sin 3x?

Question

Could anyone please explain how to figure it out? I would appreciate it. Thanks.

Answer 1

A lot.Here are some of them in (a,b) form.In fact this curve goes up and down forever.Generally all the graphs in this world stretches forever.
1)(-6.8068,9)
2)(-4.7124,9)
3)(-2.618,9)
4)(-0.5236,9)
5)(1,5708,9)
6)(3.6652,9)
7)(5,7596,9)
8)(7.854,9)

Answer 2

sin(3x) has a minimum value of (-1)
4sin(3x) has a minimum value of -4

5 - 4sin(3x) will have a maximum value of 5- (-4) = 9

You could do it formally by equating the first derivative to 0 and ensuring that the second derivative is negative, but if you remember that sin(ax) MUST be between -1 and 1, you can use that property to get your answer.

Answer 3

Max value of 4 sin 3x = 4
Min value of 4 sin 3x = - 4

Max. value of y = 5 - 4 sin 3x = 5 + 4 = 9

Answer 4

i think
max sin3x=1, -1
so =>
5-4*1=5-4=1
5-4*(-1)=5+4=9
So y(max)=9

Answer 5

As in prob 50 vide the adv link presented by potential of cpmb the issue narrows right down to proving cosA/cosB+cosB/cosC+cosC/cosA>4(cossqA+c... by potential of am>gm that's shown in a similar way as in prob50 that cosA/cosB+cosB/cosC+cosC/cosA >=2(cosA+cosB+cosC). so it greater beneficial narrows right down to proving 2(cosA+cosB+cosC)>=4(cossqA+cossqB+cossq... or cosA+cosB+cosC>=2(cossqA+cossqB+cossqC) or 2(cossqA/2+cossqB/2+cossqC/2 - 3/2) >=2(cossqA+cossqB+cossqC). Now the inequality holds for any acute angled triangle the place each and each of cossqA/2-one million/2 >=cossqA subsequently proved.

Answer 6

You cant put it like that. SIN is bout to be bond to a value of degree!

Answer 7

when x=π/2+2kπ/3 (k=0,±1,±2...)
y(max)=5-4(-1)=9