Question

19-24Explain why the function is discontinuous at the given number\(a\). Sketch the graph of the function.

24. \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{2{x^2} - 5x - 3}}{{x - 3}}}&{if\;x \ne 3}\\6&{if\;x = 3}\end{array}} \right.,\,\,a = 3\)

Short answer

The limit \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) \ne f\left( 0 \right)\)so that the function is discontinuous at \(3\).

The graph of the function is shown below:

Step-by-step solution

Write the definition of continuity of a function

A function \(f\) is said to be continuous function at a number \(a\) if \(\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = f\left( a \right)\). The function \(f\) is said to be discontinuous if it is not continuous at \(a\).

Find the average velocity over time interval

Consider the function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{2{x^2} - 5x - 3}}{{x - 3}}}&{{\rm{if}}\;x \ne 3}\\6&{{\rm{if}}\;x = 3}\end{array}} \right.\).

In the given function \(f\left( 3 \right) = 6\)is a defined function.

Now find,

\(\begin{aligned}\mathop {\lim }\limits_{x \to 3} f\left( x \right) &= \mathop {\lim }\limits_{x \to 3} \frac{{2{x^2} - 5x - 3}}{{x - 3}}\\& = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {2x + 1} \right)\left( {x - 3} \right)}}{{x - 3}}\\ &= \mathop {\lim }\limits_{x \to 3} \left( {2x + 1} \right)\\ &= 7\end{aligned}\)

The above limit exists.

Since, \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) \ne f\left( 3 \right)\).

From the above results \(f\) is discontinuousat \(3\).

Draw the graph of the function