Mastering Domain, Derivatives, and Limits: A Practical Exercise Set

Understanding the domain of functions, computing their derivatives, and evaluating limits are fundamental concepts in calculus. Below is a set of exercises designed to strengthen these skills.

Exercise Set

(a) Find the domain of the following functions:

1. $f(x) = \frac{5}{x – 3} + \sqrt{x^2 – 4x + 3}$
2. $g(x) = \frac{x}{x – 7}$

(b) Compute the first derivative of the following functions:

1. $f(x) = \frac{4x^2 – x + 3}{x + 2}$
2. $g(x) = (2x – \ln x)(e^{x^2} – 3x^2)$

(c) Evaluate the following limits:

1. $\lim\limits_{x \to 2} (3x^3 + 5x^2 – 4x + 1)$
2. $\lim\limits_{x \to 5} \frac{x^2 – 7x + 10}{x – 5}$

Step-by-Step Blog Explanation

(a) Finding the Domain of Functions

The domain of a function is the set of all possible input values $x$ for which the function is defined. There are two common restrictions:

• Denominators cannot be zero, since division by zero is undefined.
• Even roots (e.g., square roots) must have non-negative radicands, ensuring real values.

Let’s analyze the first function:

$$f(x) = \frac{5}{x – 3} + \sqrt{x^2 – 4x + 3}$$

• The denominator $x – 3$ cannot be zero, so $x \neq 3$.
• The square root function $\sqrt{x^2 – 4x + 3}$ must have a non-negative argument:

  $$x^2 – 4x + 3 \geq 0$$Factoring:

  $$(x – 3)(x – 1) \geq 0$$Solving using a sign analysis, we find $x \in (-\infty, 1] \cup [3, \infty)$, but since $x \neq 3$, the domain is:

  $$(-\infty, 1] \cup (3, \infty)$$

(b) Computing First Derivatives

Differentiation helps us analyze the rate of change of functions.

For:

$$f(x) = \frac{4x^2 – x + 3}{x + 2}$$

We use the quotient rule:

$$\left( \frac{u}{v} \right)’ = \frac{u’v – uv’}{v^2}$$

where:

• $u = 4x^2 – x + 3$, so $u’ = 8x – 1$
• $v = x + 2$, so $v’ = 1$

Applying the rule:

$$f'(x) = \frac{(8x – 1)(x + 2) – (4x^2 – x + 3)(1)}{(x + 2)^2}$$

Expanding and simplifying gives the final derivative.

(c) Evaluating Limits

For:

$$\lim\limits_{x \to 5} \frac{x^2 – 7x + 10}{x – 5}$$

Factoring the numerator:

$$x^2 – 7x + 10 = (x – 5)(x – 2)$$

Cancel the common factor:

$$\lim\limits_{x \to 5} \frac{(x – 5)(x – 2)}{x – 5} = \lim\limits_{x \to 5} (x – 2)$$

Substituting $x = 5$:

$$5 – 2 = 3$$

Thus, the limit is 3.

Conclusion

By mastering domain identification, derivative computation, and limit evaluation, students build a strong foundation for calculus. Keep practicing, and these concepts will become second nature!