How do you find an equation for the function #f'(x)=sin4x# whose graph passes through the point (pi/4,-3/4)?

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Answer 1 · 2017-02-08T02:26:07

#f(x) = -1/4cos(4x) - 1#.

Use substitution to integrate. Let #u = 4x#. Then #du = 4dx# and #dx= (du)/4#.

Integrate both sides.

#intf'(x) = 1/4intsinu du#

#f(x) = -1/4cosu + C -> "because" intsinxdx = -cosx#

#f(x) = -1/4cos(4x) + C#

Now find the value of #C#. When #x= pi/4#, #y = -3/4#, therefore:

#-3/4 = -1/4cos(4(pi/4)) + C#

#-3/4 = -1/4cos(pi) +C#

#-3/4 = 1/4 +C#

#-3/4 - 1/4 = C#

#C = -1#

Therefore, #f(x) = -1/4cos(4x) - 1#.

Hopefully this helps!