Let $$ f$$ be a differentiable function satisfying $$f(x)=1-2x+\int_{0}^{x} e^{(x-t)}f(t)dt, x \in \mathbb{R}$$ and let $$g(x)=\int_{0}^{x} (f(t)+2)^{15}(t-4)^{6}(t+12)^{17}dt, x \in \mathbb{R}.$$ If p and q are respectively the points of local minima and local maxima of g, then the value of $$\mid p+q \mid $$ is equal to _________

Find $$f(x)$$

The integral equation is $$f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) dt$$.

Let $$I(x) = \int_0^x e^{-t} f(t) dt$$. Then $$f(x) = 1 - 2x + e^x I(x)$$.

Differentiating both sides:

$$f'(x) = -2 + e^x I(x) + e^x (e^{-x} f(x))$$

Substitute $$e^x I(x) = f(x) - (1 - 2x)$$:

$$f'(x) = -2 + f(x) - 1 + 2x + f(x) \implies f'(x) - 2f(x) = 2x - 3$$.

Solving this linear ODE: $$f(x) = -x + 1 + Ce^{2x}$$. Using $$f(0)=1$$, we find $$C=0$$.

So, $$f(x) = 1 - x$$.

Finding critical points of $$g(x)$$

By Leibniz Rule: $$g'(x) = (f(x) + 2)^{15}(x - 4)^6(x + 12)^{17}$$.

Substitute $$f(x) = 1-x$$:

$$g'(x) = (1 - x + 2)^{15}(x - 4)^6(x + 12)^{17} = (3 - x)^{15}(x - 4)^6(x + 12)^{17}$$.

$$g'(x) = -(x - 3)^{15}(x - 4)^6(x + 12)^{17}$$.

Critical points are $$x = -12, 3, 4$$.

• At $$x = -12$$: $$g'(x)$$ changes from $$+$$ to $$-$$ (Local Maxima). So, $$q = -12$$.

• At $$x = 3$$: $$g'(x)$$ changes from $$-$$ to $$+$$ (Local Minima). So, $$p = 3$$.

• At $$x = 4$$: The power is even ($$6$$), so the sign doesn't change. Neither max nor min.

Final Answer: $$|p + q| = |3 + (-12)| = |-9| = \mathbf{9}$$.