MathLab with Mr. Ahmed

MathLab with Mr. Ahmed is a math learning website by Ahmed Amasha, a mathematics teacher teaching the American curriculum. Explore Algebra 2, Precalculus, Calculus, and SAT and PSAT Math Prep through lessons, focused practice and printable resources.

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Limits and Their Properties

1.1 A Preview of Calculus

Focus on the tangent-line problem, the area problem, instantaneous versus average change.

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Ideas in this lesson

01The tangent-line problemWorked steps · five questions02The area problemWorked steps · five questions03Instantaneous versus average changeWorked steps · five questions
Idea 1 of 3

The tangent-line problem

Secant slopes describe average change. Their limiting value, when it exists, defines tangent slope.

Worked example 1

For f(x)=x2f(x) = x^{2}, find the secant slope from x=2x = 2 to x=3x = 3.
  1. m=(x+h)2−x2h=2x+hm = \dfrac{(x + h)^{2} - x^{2}}{h} = 2x + h
  2. m=2(2)+1=5m = 2(2) + 1 = 5
  3. f′(2)=lim⁡h→0(2(2)+h)=4\displaystyle f'(2) = \lim\limits_{h\to 0} (2(2) + h)= 4

Answer: 55

Worked example 2

For f(x)=x2f(x) = x^{2}, find the secant slope from x=3x = 3 to x=3.1x = 3.1.
  1. m=(x+h)2−x2h=2x+hm = \dfrac{(x + h)^{2} - x^{2}}{h} = 2x + h
  2. m=2(3)+0.1=6110≈6.1m = 2(3) + 0.1 = \dfrac{61}{10} \approx 6.1
  3. f′(3)=lim⁡h→0(2(3)+h)=6\displaystyle f'(3) = \lim\limits_{h\to 0} (2(3) + h)= 6

Answer: 6110≈6.1\dfrac{61}{10} \approx 6.1

Idea 2 of 3

The area problem

Rectangles approximate curved regions. Refining a partition connects the approximation to an integral.

Worked example 1

Estimate the area below f(x)=xf(x) = x on [0,2][0, 2] using one right-endpoint rectangle.
Coordinate graph0.511.520.511.52xy
  1. Interval width =2= 2
  2. Ar=2⋅2=4A_{r} = 2 \cdot 2 = 4

Answer: 44

Worked example 2

Estimate the area below f(x)=xf(x) = x on [0,3][0, 3] using two midpoint rectangles.
Coordinate graph0.511.522.530.511.522.53xy
  1. Interval width =3= 3
  2. AAₘ =32(0.75+2.25)=92≈4.5= \dfrac{3}{2}(0.75 + 2.25) = \dfrac{9}{2} \approx 4.5

Answer: 92≈4.5\dfrac{9}{2} \approx 4.5

Idea 3 of 3

Instantaneous versus average change

Average change uses an interval; instantaneous change uses a derivative at a point.

Worked example 1

For f(x)=x2f(x) = x^{2}, find the average rate of change on [0,2][0, 2].
  1. xx0022
    f(x)f(x)0044
  2. m=4−02−0m = \dfrac{4 - 0}{2 - 0}
  3. m=2m = 2

Answer: 22

Worked example 2

For f(x)=x2f(x) = x^{2}, find the average rate of change on [1,3][1, 3].
  1. xx1133
    f(x)f(x)1199
  2. m=9−13−1m = \dfrac{9 - 1}{3 - 1}
  3. m=4m = 4

Answer: 44

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §1.1,p1.1, p. 4242. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.