Calculus

Using the knowledge that we know from our sections on limits, continuity, first derivatives, and second derivatives, we are able to proceed with graphing functions. The process that we undergo to do so is referred to as curve sketching.

The following factors are crucial when sketching a graph:

• Discontinuities/limitations in the domain
  • ex. \( f(x) = \sqrt{x} \) has the restriction that \( x \geq 0 \)

• Intercepts on the x-axis and y-axis
  • ex. \( f(x) = x^2 -1 \) has the following \(x \) and \( y \) intercepts:
    • \( f(0) = 0-1 = -1 \)                                                                       y-intercept is \( y = -1 \)
    • \( 0 = x^2 - 1 \Longrightarrow 1 = x^2 \Longrightarrow \pm \sqrt{1} = x \)                                       x-intercepts are \( x = -1, 1 \)

• Vertical and/or horizontal asymptotes
  • ex. \( f(x) = \frac{1}{x} \)
    • Looking at the denominator, we can deduce that there is a vertical asymptote at \( x = 0 \)
    • \( \lim_{x\to\infty} \frac{1}{x} = 0 \) ; there is a horizontal asymptote at \( y = 0 \) 

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  Critical points

  • ex. \( f(x) = x^3 \)
    • \( f'(x) = 3x^2 \Longrightarrow 0 = 3x^2 \Longrightarrow x = 0 \) is a critical point of \( f(x) \)

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  Points of inflection

  • ex. \( f(x) = 2x^4 + x^3 \)

    • \( f'(x) = 8x^3 + 3x^2 \)

    • \( f''(x) = 24x^2 + 6x \Longrightarrow 0 = 6x( 4x + 1) \Longrightarrow x = 0, -\frac{1}{4} \) are potential points of inflection of \( f(x) \)

Each of these factors will aid us in determining the graph's general shape, direction and behaviour.

Every function is different, and an understanding of each factor is very important. We will consider two examples and view the approach in sketching their curves. 

Consider \( f(x) = (x-1)^2(x^2 - 16) \) and \( g(x) = \frac{x^2 - x - 2}{x+ 1}\). Use curve sketching methods to draw both of their graphs on the Cartesian Plane. 

See the videos below for the solutions: