Key Concepts and Formulas

• Differentiability: A function is differentiable at a point if its derivative exists at that point. For a piecewise function, differentiability at the point where the definition changes requires:
  • Continuity at that point.
  • Equality of the left-hand derivative (LHD) and the right-hand derivative (RHD) at that point.
• Left-Hand Derivative (LHD): ${f'_{-}}(a) = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$
• Right-Hand Derivative (RHD): ${f'_{ + }}(a) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$
• Composition of Functions (gof)(x): This is defined as $g(f(x))$. When evaluating (gof)(x), we first find $f(x)$, and then use that result as the input for $g(x)$.
• Absolute Value Function: |x| = \left\{ {\matrix{ x & , & {x \ge 0} \cr -x & , & {x < 0} \cr } } \right.

Step-by-Step Solution

Step 1: Define the composite function (gof)(x). We need to find $g(f(x))$. The definition of $f(x)$ changes at $x=0$, and the definition of $g(x)$ also changes at $x=0$. Therefore, we need to consider cases based on the value of $f(x)$.

• Case 1: $f(x) < 0$ If $x < 0$, then $f(x) = -|x+3|$. Since $x < 0$, $x+3$ can be positive or negative.

  • If $-3 \le x < 0$, then $x+3 \ge 0$, so $|x+3| = x+3$. Thus, $f(x) = -(x+3) = -x-3$. In this range, $-3 \le x < 0$, so $0 \le -x \le 3$, and $-3 \le -x-3 < 0$. So $f(x) < 0$.
  • If $x < -3$, then $x+3 < 0$, so $|x+3| = -(x+3)$. Thus, $f(x) = -(-(x+3)) = x+3$. In this range, $x < -3$, so $x+3 < 0$. Thus, $f(x) < 0$. So, for $x < 0$, $f(x) < 0$. When $f(x) < 0$, $g(f(x)) = (f(x))^2 + k_1(f(x))$. For $-3 \le x < 0$, $f(x) = -x-3$. So, $(g \circ f)(x) = (-x-3)^2 + k_1(-x-3)$. For $x < -3$, $f(x) = x+3$. So, $(g \circ f)(x) = (x+3)^2 + k_1(x+3)$.

• Case 2: $f(x) \ge 0$ If $x \ge 0$, then $f(x) = e^x$. Since $x \ge 0$, $e^x \ge e^0 = 1$. So $f(x) \ge 0$. When $f(x) \ge 0$, $g(f(x)) = 4(f(x)) + k_2$. For $x \ge 0$, $f(x) = e^x$. So, $(g \circ f)(x) = 4e^x + k_2$.

Combining these, we have: (g \circ f)(x) = \left\{ {\matrix{ {(x+3)^2 + k_1(x+3)} & , & {x < -3} \cr {(-x-3)^2 + k_1(-x-3)} & , & {-3 \le x < 0} \cr {4{e^x} + {k_2}} & , & {x \ge 0} \cr } } \right.

Step 2: Ensure continuity of (gof)(x) at x = 0. For (gof) to be differentiable at $x=0$, it must first be continuous at $x=0$. We need $\lim_{x \to 0^-} (g \circ f)(x) = \lim_{x \to 0^+} (g \circ f)(x) = (g \circ f)(0)$.

• Left-hand limit at x = 0: As $x \to 0^-$, we use the definition for $-3 \le x < 0$. $\lim_{x \to 0^-} (g \circ f)(x) = \lim_{x \to 0^-} ((-x-3)^2 + k_1(-x-3))$ $= (-0-3)^2 + k_1(-0-3) = (-3)^2 + k_1(-3) = 9 - 3k_1$

• Right-hand limit at x = 0: As $x \to 0^+$, we use the definition for $x \ge 0$. $\lim_{x \to 0^+} (g \circ f)(x) = \lim_{x \to 0^+} (4e^x + k_2)$ $= 4e^0 + k_2 = 4(1) + k_2 = 4 + k_2$

• Function value at x = 0: $(g \circ f)(0) = 4e^0 + k_2 = 4 + k_2$

For continuity, $9 - 3k_1 = 4 + k_2$. $5 - 3k_1 = k_2 \quad (*)$

Step 3: Ensure differentiability of (gof)(x) at x = 0 by equating LHD and RHD. We need to find the derivatives of the relevant pieces of $(g \circ f)(x)$ and evaluate them at $x=0$.

• Right-hand derivative (RHD) at x = 0: For $x \ge 0$, $(g \circ f)(x) = 4e^x + k_2$. The derivative is ${d \over {dx}}(4e^x + k_2) = 4e^x$. The RHD at $x=0$ is ${f'_{ + }}(0) = 4e^0 = 4$.

• Left-hand derivative (LHD) at x = 0: For $-3 \le x < 0$, $(g \circ f)(x) = (-x-3)^2 + k_1(-x-3)$. Let $u = -x-3$. Then $du/dx = -1$. $(g \circ f)(x) = u^2 + k_1u$. Using the chain rule, ${d \over {dx}}(u^2 + k_1u) = (2u + k_1) \frac{du}{dx}$ $= (2(-x-3) + k_1)(-1)$ $= (-2x - 6 + k_1)(-1)$ $= 2x + 6 - k_1$ The LHD at $x=0$ is ${f'_{-}}(0) = \lim_{x \to 0^-} (2x + 6 - k_1)$ $= 2(0) + 6 - k_1 = 6 - k_1$

For differentiability, LHD = RHD: $6 - k_1 = 4$ $k_1 = 2$

Step 4: Find the value of k2 using the continuity equation. Substitute $k_1 = 2$ into the continuity equation $(*)$: $k_2 = 5 - 3k_1$ $k_2 = 5 - 3(2)$ $k_2 = 5 - 6$ $k_2 = -1$

Step 5: Evaluate (gof)(-4) and (gof)(4).

• Evaluate (gof)(-4): Since $-4 < 0$, we use the definition of $f(x)$ for $x < 0$. $f(-4) = -|-4 + 3| = -|-1| = -(1) = -1$ Now we evaluate $g(f(-4)) = g(-1)$. Since $-1 < 0$, we use the definition of $g(x)$ for $x < 0$. $g(-1) = (-1)^2 + k_1(-1)$ $= 1 + (2)(-1)$ $= 1 - 2 = -1$ So, $(g \circ f)(-4) = -1$.

• Evaluate (gof)(4): Since $4 \ge 0$, we use the definition of $f(x)$ for $x \ge 0$. $f(4) = e^4$ Now we evaluate $g(f(4)) = g(e^4)$. Since $e^4 \ge 0$, we use the definition of $g(x)$ for $x \ge 0$. $g(e^4) = 4(e^4) + k_2$ $= 4e^4 + (-1)$ $= 4e^4 - 1$ So, $(g \circ f)(4) = 4e^4 - 1$.

Step 6: Calculate (gof)(-4) + (gof)(4). $(g \circ f)(-4) + (g \circ f)(4) = (-1) + (4e^4 - 1)$ $= 4e^4 - 2$ $= 2(2e^4 - 1)$

Alternative approach for Step 5 and 6 using the derived (gof)(x) definition: We found $k_1=2$ and $k_2=-1$. The definition of $(g \circ f)(x)$ is: (g \circ f)(x) = \left\{ {\matrix{ {(x+3)^2 + 2(x+3)} & , & {x < -3} \cr {(-x-3)^2 + 2(-x-3)} & , & {-3 \le x < 0} \cr {4{e^x} - 1} & , & {x \ge 0} \cr } } \right.

• Evaluate (gof)(-4): Since $-4 < -3$, we use the first case. $(g \circ f)(-4) = (-4+3)^2 + 2(-4+3)$ $= (-1)^2 + 2(-1)$ $= 1 - 2 = -1$

• Evaluate (gof)(4): Since $4 \ge 0$, we use the third case. $(g \circ f)(4) = 4e^4 - 1$

• Calculate (gof)(-4) + (gof)(4): $(g \circ f)(-4) + (g \circ f)(4) = (-1) + (4e^4 - 1)$ $= 4e^4 - 2$ $= 2(2e^4 - 1)$

Common Mistakes & Tips

• Incorrectly applying the absolute value: Be careful when evaluating $|x+3|$ for different ranges of $x$.
• Mixing up the definitions of $f(x)$ and $g(x)$ when composing: Ensure you are using the correct function for the input value. For $(g \circ f)(x)$, first evaluate $f(x)$, and then use that output as the input for $g(x)$.
• Forgetting the continuity condition: Differentiability implies continuity. Always check for continuity first, as it provides necessary equations for the constants.
• Chain Rule Errors: When differentiating composite functions, especially piecewise ones, ensure the chain rule is applied correctly. For $(g \circ f)(x)$, the derivative is $g'(f(x)) \cdot f'(x)$. However, it's often easier to derive the explicit form of $(g \circ f)(x)$ first and then differentiate.

Summary

To solve this problem, we first defined the composite function $(g \circ f)(x)$. For the function to be differentiable at $x=0$, it must be continuous and have equal left-hand and right-hand derivatives at $x=0$. We used the continuity condition to establish a relationship between the constants $k_1$ and $k_2$. Then, we calculated the left-hand and right-hand derivatives of $(g \circ f)(x)$ at $x=0$ and equated them to find the value of $k_1$. Using the value of $k_1$, we found the value of $k_2$. Finally, we substituted these values back into the expression for $(g \circ f)(x)$ and calculated $(g \circ f)(-4) + (g \circ f)(4)$.

The final answer is $4e^4 - 2$, which is equal to $2(2e^4 - 1)$.

The final answer is \boxed{2(2{e^4} - 1)}.