Point of discontinuity

Point of discontinuity

Lessons

• "point of discontinuity" exists when the numerator and denominator have a factor in common.
i.e. (x)=(3x8)(x+5)(2x7)(x+5)(4x+9)(3x+8)(2x7)\left( x \right) = \frac{{ - \left( {3x - 8} \right)\left( {x + 5} \right)\left( {2x - 7} \right)}}{{\left( {x + 5} \right)\left( {4x + 9} \right)\left( {3x + 8} \right)\left( {2x - 7} \right)}} ; points of discontinuity exist at x=5x = - 5 and x=72x = \frac{7}{2} .
• To determine the coordinates of the point of discontinuity:
1) Factor both the numerator and denominator.
2) Simplify the rational expression by cancelling the common factors.
3) Substitute the non-permissible values of x into the simplified rational expression to obtain the corresponding values for the y-coordinate.
  • 1.
    Investigating How a Point of Discontinuity Appears on a Graph
    Sketch and compare the following two functions:
    i) f(x)=2x+5f\left( x \right) = 2x + 5
    ii) g(x)=2x2+11x+15x+3g\left( x \right) = \frac{{2{x^2} + 11x + 15}}{{x + 3}}

  • 2.
    Sketching Rational Functions Incorporating Asymptotes and Points of Discontinuity
    Sketch the rational function: f(x)=2x27x+52x311x2+19x10f\left( x \right) = \frac{{2{x^2} - 7x + 5}}{{2{x^3} - 11{x^2} + 19x - 10}}