What is the maximum value of y for the equation y=1+3sinx?

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Answer 1 · 2018-03-26T14:34:46

Maximum value of #y# for the equation #y=1+3sinx# is #4#.

As maximum value for #sinx# is #1#, maximumvalue for #y=1+3sinx# is #1+3*1=4#.

Let us still try it out using calculas. Maxima appears when #(dy)/(dx)=0# and #(d^2y)/(dx^2)<0#

As #y=1+3sinx#, #(dy)/(dx)=3cosx#, which is #0#, when #x=pi/2#.

and #(d^2y)/(dx^2)=-3sinx# and at #x=pi/2#,

#(d^2y)/(dx^2)# at #x=pi/2# is #-3*1=-3#

as such at #x=pi/2#, #(dy)/(dx)=0# and #(d^2y)/(dx^2)<0#, we have a local maxima, which is

#y=1+3sin(pi/2)=1+3*1=4#