Math
posted by Vanessa .
Find the function f, given that the slope of the tangent line at any point (x,f(x)) is f '(x) and that the graph of f passes through the given point.
f '(x)=6(2x7)^5 at (4, 3/2)

f=INT f= (2x7)^6 + C
now find C
3/2 = (2*47)^6 + C
C=1/2 
if dy/dx = 6(2x7)^5
y = (6/6) (2x7)^6 (1/2) + c
y = (1/2)(2x7)^6 + c
but (4 , 3/2) lies on it, so
3/2 = (1/2)(1)^6 + c
3/2 = 1/2 + c
c = 1
f(x) = (1/2)(2x7)^6 + 1