Key Concepts and Formulas

• Continuity of a function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. For piecewise functions, this condition must hold at the points where the definition of the function changes.
• Differentiability of a function: A function $f(x)$ is differentiable at a point $x=c$ if the left-hand derivative and the right-hand derivative at $x=c$ are equal. That is, $\lim_{h \to 0^-} \frac{f(c+h)-f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h}$.
• Derivative of common functions:
  • The derivative of $e^x$ is $e^x$.
  • The derivative of $e^{-x}$ is $-e^{-x}$.
  • The derivative of $x^n$ is $nx^{n-1}$.
  • The derivative of $cx^2$ is $2cx$.
  • The derivative of $ax^2$ is $2ax$.

Step-by-Step Solution

Step 1: Analyze the given function and the conditions. The function $f(x)$ is defined piecewise: f\left( x \right) = \left\{ {\matrix{ {a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr {c{x^2},} & {1 \le x \le 3} \cr {a{x^2} + 2cx,} & {3 < x \le 4} \cr } } \right. We are given that the function is continuous for some $a, b, c \in \mathbb{R}$. This implies continuity at the points where the definition changes, i.e., at $x=1$ and $x=3$. We are also given the condition $f'(0) + f'(2) = e$.

Step 2: Apply the continuity condition at $x=1$. For continuity at $x=1$, we must have $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$. The left-hand limit is $\lim_{x \to 1^-} (a{e^x} + b{e^{ - x}}) = ae^1 + be^{-1} = ae + \frac{b}{e}$. The right-hand limit is $\lim_{x \to 1^+} (c{x^2}) = c(1)^2 = c$. Equating these, we get: $ae + \frac{b}{e} = c \quad (*)$ We can rearrange this to express $b$ in terms of $a$ and $c$: $b = e(c - ae) = ce - a{e^2}$.

Step 3: Apply the continuity condition at $x=3$. For continuity at $x=3$, we must have $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)$. The left-hand limit is $\lim_{x \to 3^-} (c{x^2}) = c(3)^2 = 9c$. The right-hand limit is $\lim_{x \to 3^+} (a{x^2} + 2cx) = a(3)^2 + 2c(3) = 9a + 6c$. Equating these, we get: $9c = 9a + 6c$ $3c = 9a$ $c = 3a \quad (**)$

Step 4: Find the derivatives of the relevant parts of the function. We need $f'(0)$ and $f'(2)$. For $x \in (-1, 1)$, $f(x) = ax{e^x} + b{e^{ - x}}$. The derivative is $f'(x) = ae^x - be^{-x}$. For $x \in (1, 3)$, $f(x) = cx^2$. The derivative is $f'(x) = 2cx$. For $x \in (3, 4)$, $f(x) = ax^2 + 2cx$. The derivative is $f'(x) = 2ax + 2c$.

Step 5: Calculate $f'(0)$ and $f'(2)$. $f'(0)$ is calculated using the derivative for $x \in (-1, 1)$: $f'(0) = ae^0 - be^{-0} = a - b$ $f'(2)$ is calculated using the derivative for $x \in (1, 3)$: $f'(2) = 2c(2) = 4c$

Step 6: Use the given condition $f'(0) + f'(2) = e$. Substitute the calculated values of $f'(0)$ and $f'(2)$: $(a - b) + 4c = e \quad (***)$

Step 7: Substitute the relations from continuity conditions into the derivative condition. We have three equations:

1. $b = ce - a{e^2}$ (from Step 2)
2. $c = 3a$ (from Step 3)
3. $a - b + 4c = e$ (from Step 6)

Substitute equation (2) into equation (1): $b = (3a)e - a{e^2} = 3ae - a{e^2}$

Now substitute the expressions for $b$ and $c$ in terms of $a$ into equation (3): $a - (3ae - a{e^2}) + 4(3a) = e$ $a - 3ae + a{e^2} + 12a = e$

Step 8: Solve for $a$. Combine the terms involving $a$: $a(1 - 3e + e^2 + 12) = e$ $a(e^2 - 3e + 13) = e$

Now, isolate $a$: $a = \frac{e}{e^2 - 3e + 13}$

Common Mistakes & Tips

• Algebraic Errors: Be very careful with algebraic manipulations, especially when dealing with exponential terms and signs. Errors in substituting or rearranging equations are common.
• Incorrect Derivative Application: Ensure you use the correct form of the derivative for the given interval of $x$. For example, $f'(0)$ must use the derivative of $a{e^x} + b{e^{ - x}}$, not $c{x^2}$.
• Misinterpreting Continuity: Remember that continuity at the boundary points requires the limits from both sides to be equal to the function's value at that point. For piecewise functions, this means equating the expressions defining the function on either side of the boundary.

Summary

The problem requires us to find the value of $a$ given a piecewise function and conditions for continuity and a specific relationship between its derivatives. We first used the continuity conditions at $x=1$ and $x=3$ to establish relationships between $a$, $b$, and $c$. Specifically, continuity at $x=1$ gave us $ae + b/e = c$, and continuity at $x=3$ yielded $c=3a$. We then calculated the derivatives of the relevant function pieces and used the given condition $f'(0) + f'(2) = e$ to form a third equation. By substituting the relationships derived from continuity into this equation, we obtained a single equation in terms of $a$, which we solved to find the value of $a$.

The final answer is \boxed{\frac{e}{{{e^2} - 3e + 13}}}.