Since f'(x) = g'(x), the only difference between f(x) and g(x) is their constant term.

Let h(x) be a function with no constant term.
Let h'(x) = f'(x) = g'(x)

Then,

f(x) = ∫ h'(x) dx
f(x) = h(x) + c , c is a constant

g(x) = ∫ h'(x) dx
g(x) = h(x) + d, d is a constant


Sub (3,6) ,

f(3) = h(3) + c
6 = h(3) + c ①

Sub (3,-1) ,

g(3) = h(3) + d
-1 = h(3) + d ②

① - ②

6 - (-1) = h(3) + c - (h(3) + d)
7 = c - d
d = c - 7


So,

g(x)
= h(x) + d
= h(x) + c - 7
= f(x) - 7

Example :

f(x) = 3x - 3
When x = 3, f(3) = 3(3) - 3 = 6


f'(x) = 3x


g(x) = 3x - 10
When x = 3, f(3) = 3(3) - 10 = -1


g'(x) = 3x = f'(x)

Alternative working :

∫ f'(x) dx = f(x) + c, c is a constant

(The indefinite integral of f'(x) does not have the constant term that f(x) has ,so the c is added to equate them)

Example : If f(x) = 2x² + 5, then f'(x) = 4x

∫ f'(x) dx
= ∫ 4x dx
= 2x² + a , where a is a constant
= 2x² + 5 + a - 5
= f(x) + (a - 5)
= f(x) + c, where c = a - 5

Similarly,
∫ g'(x) dx = g(x) + d , d is a constant


Since f'(x) = g'(x) for all real values of x,

∫ f'(x) dx = ∫ g'(x) dx

f(x) + c = g(x) + d
g(x) = f(x) + c - d

Since (3,6) and (3,-1) lie on the graphs of y = f(x) and y = g(x) respectively,

f(3) = 6 and g(3) = -1

Then, when x = 3,
g(3) = f(3) + c - d
-1 = 6 + c - d
c - d = -7

So g(x) = f(x) - 7