Is the graph of f (x) = (X^2 + x)/x continuous on the interval [-4, 4]?

Feb 9, 2017

see below

Explanation:

To show that the function $f \left(x\right) = \frac{{x}^{2} + x}{x}$ is continuous on [-4,4] we need to do the following:

1. Show that ${\lim}_{x \to - {4}^{+}} \frac{{x}^{2} + x}{x} = f \left(- 4\right)$

2. Show that ${\lim}_{x \to {4}^{-}} \frac{{x}^{2} + x}{x} = f \left(4\right)$

That is,

1. ${\lim}_{x \to - {4}^{+}} \frac{{x}^{2} + x}{x} = {\lim}_{x \to - {4}^{+}} \frac{x \left(x + 1\right)}{x}$

$= {\lim}_{x \to - {4}^{+}} \frac{\cancel{x} \left(x + 1\right)}{\cancel{x}} = {\lim}_{x \to - {4}^{+}} x + 1 = - 3$

$f \left(- 4\right) = \frac{16 - 4}{-} 4 = \frac{12}{-} 4 = - 3$

Since ${\lim}_{x \to - {4}^{+}} \frac{{x}^{2} + x}{x} = f \left(- 4\right) = - 3$, f is continuous from the right at x = - 4.

2.${\lim}_{x \to {4}^{-}} \frac{{x}^{2} + x}{x} = {\lim}_{x \to {4}^{-}} \frac{x \left(x + 1\right)}{x}$

$= {\lim}_{x \to {4}^{-}} \frac{\cancel{x} \left(x + 1\right)}{\cancel{x}} = {\lim}_{x \to {4}^{-}} x + 1 = 5$

$f \left(4\right) = \frac{16 + 4}{4} = \frac{20}{4} = 5$

Since ${\lim}_{x \to {4}^{-}} \frac{{x}^{2} + x}{x} = f \left(4\right)$, f is continuous from the left at 4.

Since f is continuous from the left at x = 4and continuous from the right at x = -4, hence f is continuous on the interval [-4,4].