The Lumenary · Grades 9–12

Introductory calculus

Name: ______________________

Date: ______________________

You will learn to

Estimate limits from tables and graphs, compute simple limits algebraically, and explain why a limit need not equal the value of the function
Define the derivative as a limit of difference quotients, and differentiate polynomials from the definition and with the power rule
Use derivatives to find tangent lines, rates of change and extrema — knowing that $f'(a) = 0$ does not guarantee a maximum or minimum
Approximate areas with Riemann sums, find the area under $y = x^2$ as a limit, and interpret integrals as signed areas
Evaluate integrals with antiderivatives using the fundamental theorem of calculus

1.
Find $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ .
2.
Let $f(x) = 4x^3 - 5x + 2$ . Compute $f'(1)$ .
3.
The tangent line to $y = x^2 - 4x$ at $x = 3$ has equation $y = mx + c$ . Find $c$ .
4.
For $f(x) = x^3 - 12x$ , which statement is correct?
- A.Local maximum $16$ at $x = -2$ , local minimum $-16$ at $x = 2$
- B.Local minimum $16$ at $x = -2$ , local maximum $-16$ at $x = 2$
- C.Turning points at $x = \pm 12$
- D.No turning points, since $x^3$ has none
5.
Which function is a counterexample to the claim "if $f'(0) = 0$ then $f$ has a local maximum or minimum at $0$ "?
- A. $f(x) = x^3$
- B. $f(x) = x^2$
- C. $f(x) = |x|$
- D. $f(x) = x^4$
6.
Compute $\displaystyle\int_0^3 (x^2 - 2x)\,dx$ .
7.
In the proof of Fermat's theorem, $f$ has a local maximum at $a$ . Why is $\dfrac{f(a + h) - f(a)}{h} \ge 0$ for small negative $h$ ?
- A.The numerator is $\le 0$ (maximum) and the denominator is $< 0$
- B.Because $f$ is increasing to the left of $a$
- C.Because $f'(a) = 0$
- D.Because difference quotients are always non-negative
8.
In Lean, with f := fun x : ℝ => (x ^ 2 - 1) / (x - 1), the lesson proves both Tendsto f (𝓝[≠] 1) (𝓝 2) and f 1 = 0. What is the right conclusion?
- A.Lean defines $x / 0 = 0$ , so f 1 = 0; the limit ignores the value at $1$ , so there is no contradiction
- B.Lean is inconsistent, because the limit should equal the value
- C.The limit is really $0$ , and the lesson's proof has a mistake
- D.𝓝[≠] 1 means that $x$ is allowed to equal $1$
9.
Write right_sum(f, a, b, n) returning the right Riemann sum $\sum_{k=1}^{n} f\!\left(a + k\,\tfrac{b - a}{n}\right)\tfrac{b - a}{n}$ . Use Fraction so that the sums are exact when a, b are fractions and f is a polynomial. Then check Theorem 7: for $f(x) = x^2$ on $[0, 1]$ it should equal $\tfrac{(n + 1)(2n + 1)}{6n^2}$ .
Computational — Practice

Answer key — Introductory calculus

1.
For $x \ne 3$ , $\dfrac{x^2 - 9}{x - 3} = \dfrac{(x - 3)(x + 3)}{x - 3} = x + 3$ . As $x \to 3$ this tends to $6$ . The function is undefined at $x = 3$ , but the limit is $6$ .
2.
$f'(x) = 4 \cdot 3x^2 - 5 + 0 = 12x^2 - 5$ , so $f'(1) = 12 - 5 = 7$ .
3.
$f(3) = 9 - 12 = -3$ and $f'(x) = 2x - 4$ , so $f'(3) = 2$ . The tangent is $y = -3 + 2(x - 3) = 2x - 9$ , so $c = -9$ .
4.
$f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$ is zero at $x = \pm 2$ . It is positive for $x < -2$ , negative on $(-2, 2)$ and positive for $x > 2$ . So $f$ rises to a local maximum $f(-2) = -8 + 24 = 16$ and falls to a local minimum $f(2) = 8 - 24 = -16$ .
5.
$f(x) = x^3$ : $f'(x) = 3x^2$ , so $f'(0) = 0$ , yet $f$ is increasing — positive to the right of $0$ and negative to the left — so $0$ is not an extremum. Fermat's theorem (extremum $\Rightarrow f' = 0$ ) has a false converse.
6.
By the fundamental theorem, $\int_0^3 (x^2 - 2x)\,dx = \big[\tfrac{x^3}{3} - x^2\big]_0^3 = (9 - 9) - 0 = 0$ . The graph is below the axis on $(0, 2)$ , contributing $\tfrac83 - 4 = -\tfrac43$ , and above it on $(2, 3)$ , contributing $+\tfrac43$ . The signed areas cancel; the total geometric area is $\tfrac83$ .
7.
At a local maximum, $f(a + h) - f(a) \le 0$ for all small $h$ . When $h < 0$ , dividing a non-positive number by a negative number gives a non-negative result, so the quotient is $\ge 0$ and its limit $f'(a) \ge 0$ . For $h > 0$ the quotient is $\le 0$ , so $f'(a) \le 0$ . Hence $f'(a) = 0$ .
8.
Lean's convention $x / 0 = 0$ makes f 1 = (1 - 1) / (1 - 1) = 0 / 0 = 0. The limit is taken along 𝓝[≠] 1, i.e. over $x \ne 1$ , where $f(x) = x + 1 \to 2$ . A limit never depends on the value at the point, so both statements are true and consistent; they show that $f$ (with Lean's value) is discontinuous at $1$ .
9.
```
from fractions import Fraction

def right_sum(f, a, b, n):
    width = Fraction(b - a) / n
    total = Fraction(0)
    for k in range(1, n + 1):
        total += f(a + k * width) * width
    return total
```
For $f(x) = x^2$ on $[0, 1]$ the sum is $\frac{1^2 + 2^2 + \dots + n^2}{n^3} = \frac{(n + 1)(2n + 1)}{6n^2}$ , which the loop confirms exactly for $n < 60$ — evidence that agrees with the proof by induction. As $n$ grows these values approach $\frac13$ .