How do you find an equation for the function #f'(x)=cos(x/2)# whose graph passes through the point (0,3)?

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Answer 1 · 2018-02-14T18:32:38

Integrate both sides of the given equation and then use the point, #(0,3)# to solve for the constant of integration.

Given:

#f'(x)=cos(x/2)#

integrate both sides:

#int f'(x) dx = int cos(x/2)dx#

#f(x) = 1/(1/2)sin(x/2) +C #

#f(x) = 2sin(x/2) +C #

Use the point #(0,3)# to solve for C:

#f(0) = 2sin(0/2)+C= 3#

#C = 3#

#f(x) = 2sin(x/2) + 3#