Let f be a differentiable function satisfying f(x) = 1 - 2x + ∫₀ˣ e⁽ˣ⁻ᵗ⁾ f(t) dt, x ∈ R and let g(x)

Q: Wait, why can't I just differentiate the integral directly? The Leibniz rule seems complicated.

Great question! You *can* differentiate directly, but if you don't separate the $e^x$ term first, you'll end up with a messy expression involving both $x$ and $t$ inside the derivative. By pulling $e^x$ out, you make the integral a function of $x$ multiplied by $e^x$, which makes the Product Rule application much cleaner.

Q: Why does the sign of $g'(x)$ not change at $x=4$?

This is a classic trap! Look at the factor $(x-4)^6$. Because the exponent $6$ is an even number, the term $(x-4)^6$ is always non-negative, regardless of whether $x$ is slightly greater than or less than $4$. Therefore, the sign of the entire expression $g'(x)$ is determined by the other factors, not by this one. Always check the parity of the exponent!

Q: How did we get $f(0)=1$ so easily?

Look at the original equation: $f(x) = 1 - 2x + \int_{0}^{x} e^{(x-t)} f(t) dt$. If you set $x=0$, the integral becomes $\int_{0}^{0} \dots dt$. By the fundamental properties of definite integrals, the integral from any point to itself is always zero. Thus, $f(0) = 1 - 2(0) + 0 = 1$. It's a beautiful simplification that saves us a lot of work!

Question

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 $|p + q|$ is equal to .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, intimidating integral equation. It feels like a locked door, but every lock has a key. Our journey begins with the equation:

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

The trap here is the $e^{(x-t)}$ term, which binds $x$ and $t$ together. By applying the laws of exponents, we rewrite this as $e^{(x-t)} = e^x \cdot e^{-t}$.

By pulling the $e^x$ term out of the integral, we transform the equation into:

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

Now that the variable $x$ is isolated, we can apply the Leibniz Rule with confidence.

### The Power of the Leibniz Rule

We differentiate both sides with respect to $x$. On the right side, we apply the Product Rule to the term $e^x \int_{0}^{x} e^{-t} f(t) dt$.

The Leibniz Rule states that the derivative of the integral $\int_{0}^{x} h(t) dt$ is simply $h(x)$. Applying this, we obtain:

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

Notice that the term $e^x \int_{0}^{x} e^{-t} f(t) dt$ reappears. This is exactly the integral part of our original equation, which we can substitute as $f(x) - 1 + 2x$.

This substitution simplifies our derivative into a beautiful, linear first-order differential equation:

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

### Solving the ODE

We are now in familiar territory. A linear ODE of the form $\frac{dy}{dx} + Py = Q$ is solved using an Integrating Factor (I.F.). Here, $P = -1$, so our I.F. is $e^{\int -1 dx} = e^{-x}$.

Multiplying our ODE by this factor, we get:

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

Using Integration by Parts, we solve this integral to find $f(x) = 1 - 2x + C e^x$. By plugging $x=0$ into the original equation, we find $f(0)=1$, which forces $C=0$. Thus, our function is simply **$f(x) = 1 - 2x$**.

### The Wavy Curve Method

Finally, we turn to $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^6 (t + 12)^{17} dt$. Differentiating this with the Leibniz Rule gives:

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

Substituting $f(x) = 1 - 2x$, we get:

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

Rearranging to keep the $x$ terms positive, we have:

$$g'(x) = -2^{15} \left(x - \frac{3}{2}\right)^{15} (x - 4)^6 (x + 12)^{17}$$

The critical points are $x = -12$, $x = 1.5$, and $x = 4$. By analyzing the signs of these factors, we find that at $x = -12$, the slope changes from negative to positive (a local minimum, $p = -12$). At $x = 1.5$, it changes from positive to negative (a local maximum, $q = 1.5$). At $x = 4$, the sign does not change because the exponent is even.

The final answer is $|-12 + 1.5| = |-10.5| = \mathbf{10.5}$.

Enter Numerical Value:

Visualized Solution

Analyze the Integral Equation

Differentiate using Leibniz Rule

Simplify to a Linear ODE

Finding Integrating Factor

Solve the ODE for $f (x)$

Apply Initial Condition

Find the Derivative $g^{'} (x)$

Identify Critical Points

Sign Analysis of $g^{'} (x)$

Identify $p$ and $q$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Power of the Leibniz Rule

Solving the ODE

The Wavy Curve Method

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