Find f if f"(theta)=sin (theta) + cos (theta), f(0)=5, and f'(0)=8?

Question

Find f if f"(theta)=sin (theta) + cos (theta), f(0)=5, and f'(0)=8

Answer 1

f if f"(theta)=sin (theta) + cos (theta), f(0)=5, and f'(0)=8

f'(x)= -cos (theta)+sin (theta) +C
0=-1+0+C
C=-1

f(x)=-sin(theta)-cos (theta)-x+C
0=-0-(-1)-0+C
C=1

f(x)=-sin(theta)-cos (theta)-x+1

Answer 2

its the antiderivative right? theta=x
F"(x)=sin(x)+cos(x)
F'(x)= -cos(x)+sin(x) + c plug in 0 for x and find c
F(x)= -sin(x)-cos(x)+cx+k after finding c, plug in 0 and find k
double check to ensure its 100% right though

Answer 3

f"(theta)=sin (theta) + cos (theta)
f '(theta) = -cos(theta) + sin(theta) and f '(0) = 8, so
-cos(theta) + sin(theta) = 8
This cannot be true, since -cos(theta) + sin(theta) <=2
Then f ' does not exist.
Then f does not exist.

Th

Answer 4

woah, that's complicated. good luck to everyone