Solving for the Differentiable Function $f(x)$

The given integral equation is:

$f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt, \quad x \in \mathbf{R}$

Rewrite the integral term:

$f(x) = 1 - 2x + e^{x} \int_{0}^{x} e^{-t} f(t) dt$

Differentiate both sides with respect to $x$. Using the product rule and the Fundamental Theorem of Calculus:

$f'(x) = -2 + \left( e^{x} \int_{0}^{x} e^{-t} f(t) dt \right) + e^{x} \left( \frac{d}{dx} \int_{0}^{x} e^{-t} f(t) dt \right)$

$f'(x) = -2 + e^{x} \int_{0}^{x} e^{-t} f(t) dt + e^{x} (e^{-x} f(x))$

$f'(x) = -2 + e^{x} \int_{0}^{x} e^{-t} f(t) dt + f(x)$

From the rewritten integral equation, we know that $e^{x} \int_{0}^{x} e^{-t} f(t) dt = f(x) - 1 + 2x$. Substitute this back:

$f'(x) = -2 + (f(x) - 1 + 2x) + f(x)$

$f'(x) = 2f(x) + 2x - 3$

This is a linear first-order differential equation: $f'(x) - 2f(x) = 2x - 3$.

The integrating factor is $e^{\int -2 dx} = e^{-2x}$.

Multiplying the equation by $e^{-2x}$ gives:

$(e^{-2x} f(x))' = (2x - 3) e^{-2x}$

Integrate both sides:

$e^{-2x} f(x) = \int (2x - 3) e^{-2x} dx$

Using integration by parts, we find:

$e^{-2x} f(x) = (-x + 1) e^{-2x} + C$

$f(x) = 1 - x + C e^{2x}$

To find $C$, use the initial condition $f(0) = 1 - 2(0) + \int_{0}^{0} \dots dt = 1$.

$f(0) = 1 - 0 + C e^{0} \implies 1 = 1 + C \implies C = 0$

Therefore, the function is $f(x) = 1 - x$.

Analyzing the Derivative of $g(x)$

The function $g(x)$ is defined as:

$g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$

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

$g(x) = \int_{0}^{x} ((1 - t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$

$g(x) = \int_{0}^{x} (3 - t)^{15} (t - 4)^{6} (t + 12)^{17} dt$

By the Fundamental Theorem of Calculus, the derivative $g'(x)$ is:

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

Identifying Local Minima and Maxima Points

Local extrema occur where $g'(x) = 0$. The critical points are the roots of $g'(x)$:

$(3 - x)^{15} (x - 4)^{6} (x + 12)^{17} = 0$

The critical points are $x = 3$, $x = 4$, and $x = -12$. Now, analyze the sign changes of $g'(x)$ around these points.

• Sign Analysis of $g'(x) = (3 - x)^{15} (x - 4)^{6} (x + 12)^{17}$:
  • The term $(x - 4)^{6}$ is always non-negative ($ \ge 0$).
  • The sign of $(3 - x)^{15}$ is determined by $(3 - x)$.
  • The sign of $(x + 12)^{17}$ is determined by $(x + 12)$.

Interval $(-\infty, -12)$: $g'(x)$ is (positive) * (positive) * (negative) = negative.

Interval $(-12, 3)$: $g'(x)$ is (positive) * (positive) * (positive) = positive.

At $x = -12$, $g'(x)$ changes from negative to positive. This indicates a local minimum. So, $p = -12$.

Interval $(3, 4)$: $g'(x)$ is (negative) * (positive) * (positive) = negative.

At $x = 3$, $g'(x)$ changes from positive to negative. This indicates a local maximum. So, $q = 3$.

Interval $(4, \infty)$: $g'(x)$ is (negative) * (positive) * (positive) = negative.

At $x = 4$, the sign of $g'(x)$ does not change (it remains negative on both sides). Thus, $x=4$ is neither a local minimum nor a maximum.

Calculating $|p + q|$

We found the point of local minimum $p = -12$ and the point of local maximum $q = 3$.

The required value is $|p + q|$:

$|p + q| = |-12 + 3| = |-9| = 9$