Understanding Continuity, Derivatives, and Function Composition: Step-by-Step Solutions

Exercise Set

(a) Find the values of $a$ for which the function is continuous at all points in its domain:

$$f(x) = \begin{cases} x^2 – 4x + 3, & x < 5 \\ ax + 1, & x \geq 5 \end{cases}$$

(b) Compute the derivative of the function:

$$f(x) = (5e^x – 2x^3)^4$$

(c) Determine the range of the following functions, assuming their domain is all $\mathbb{R}$:

1. $f(x) = -3x^2 + 6x + 9$
2. $g(x) = e^{-x} + 4$

(d) Given the functions $f(x) = 3x – 2$ and $g(x) = x^2 + 5$, determine $f(g(x))$.

Step-by-Step Solutions

(a) Ensuring Continuity

A function is continuous at a point $x = c$ if:

1. $\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x)$ (Left-hand limit equals right-hand limit)
2. The common limit equals $f(c)$.

Here, continuity at $x = 5$ means:

$$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x) = f(5)$$

Step 1: Compute $\lim\limits_{x \to 5^-} f(x)$ (using $x^2 – 4x + 3$)

Substituting $x = 5$:

$$5^2 – 4(5) + 3 = 25 – 20 + 3 = 8$$

Step 2: Compute $\lim\limits_{x \to 5^+} f(x)$ (using $ax + 1$)

Substituting $x = 5$:

$$a(5) + 1 = 5a + 1$$

For continuity:

$$8 = 5a + 1$$

Solving for $a$:

$$5a = 7 \quad \Rightarrow \quad a = \frac{7}{5}$$

(b) Computing the Derivative

The function:

$$f(x) = (5e^x – 2x^3)^4$$

We apply the chain rule, which states:

$$\frac{d}{dx} [g(h(x))] = g'(h(x)) \cdot h'(x)$$

Let $u = 5e^x – 2x^3$, so:

$$f(x) = u^4$$

Using the chain rule:

$$f'(x) = 4u^3 \cdot u’$$

Differentiate $u$:

$$u’ = 5e^x – 6x^2$$

Thus:

$$f'(x) = 4(5e^x – 2x^3)^3 \cdot (5e^x – 6x^2)$$

(c) Finding the Range

(i) $f(x) = -3x^2 + 6x + 9$

This is a quadratic function of the form $ax^2 + bx + c$. Since $a = -3$ (negative), the parabola opens downward, meaning it has a maximum value at its vertex.

The vertex occurs at:

$$x = \frac{-b}{2a} = \frac{-6}{2(-3)} = \frac{6}{6} = 1$$

Substituting $x = 1$:

$$f(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$$

Since the parabola extends to $-\infty$, the range is:

$$(-\infty, 12]$$

(ii) $g(x) = e^{-x} + 4$

Since $e^{-x}$ is always positive and approaches 0 as $x \to \infty$, its minimum value occurs at:

$$\lim\limits_{x \to \infty} e^{-x} + 4 = 0 + 4 = 4$$

Its maximum occurs at $x \to -\infty$, where $e^{-x} \to \infty$, so:

$$\lim\limits_{x \to -\infty} e^{-x} + 4 = \infty + 4 = \infty$$

Thus, the range is:

$$(4, \infty)$$

(d) Function Composition: $f(g(x))$

We substitute $g(x)$ into $f(x)$:

Given:

$$f(x) = 3x – 2, \quad g(x) = x^2 + 5$$ $$f(g(x)) = 3(x^2 + 5) – 2$$

Expanding:

$$f(g(x)) = 3x^2 + 15 – 2 = 3x^2 + 13$$

Thus, the composed function is:

$$f(g(x)) = 3x^2 + 13$$

Final Answers Summary

1. Continuity Condition: $a = \frac{7}{5}$
2. Derivative: $$f'(x) = 4(5e^x – 2x^3)^3 \cdot (5e^x – 6x^2)$$
3. Ranges:
   • $(-\infty, 12]$ for $f(x) = -3x^2 + 6x + 9$
   • $(4, \infty)$ for $g(x) = e^{-x} + 4$
4. Composition: $$f(g(x)) = 3x^2 + 13$$

Conclusion

This step-by-step approach ensures a clear understanding of continuity, differentiation, function range, and composition. These fundamental topics build the foundation for more advanced calculus applications. Keep practicing, and calculus will soon feel intuitive! 🚀