# How do you use the definition of continuity and the properties of limits to show that the function g(x) = sqrt(-x^2 + 8*x - 15) is continuous on the interval [3,5]?

Jun 15, 2015

There is no one sentence answer.

#### Explanation:

In order for $g$ to be continuous on $\left[3 , 5\right]$, the definition of continuous on a closed interval requires:

For $c$ in $\left(3 , 5\right)$, we need${\lim}_{x \rightarrow c} g \left(x\right) = g \left(c\right)$
and we also need one-sided continuity at the endpoints:
we need: ${\lim}_{x \rightarrow {3}^{+}} g \left(x\right) = g \left(3\right)$ and ${\lim}_{x \rightarrow {5}^{-}} g \left(x\right) = g \left(5\right)$

For $c$ in $\left(3 , 5\right)$, We'll use the properties of limits to evaluate the limit:

${\lim}_{x \rightarrow c} g \left(x\right) = {\lim}_{x \rightarrow c} \sqrt{- {x}^{2} + 8 x - 15}$

$= \sqrt{{\lim}_{x \rightarrow c} \left(- {x}^{2} + 8 x - 15\right)}$

$= \sqrt{{\lim}_{x \rightarrow c} \left(- {x}^{2}\right) + {\lim}_{x \rightarrow c} \left(8 x\right) - {\lim}_{x \rightarrow c} \left(15\right)}$

$= \sqrt{- {\lim}_{x \rightarrow c} \left({x}^{2}\right) + 8 {\lim}_{x \rightarrow c} \left(x\right) - {\lim}_{x \rightarrow c} \left(15\right)}$

$= \sqrt{- {\left({\lim}_{x \rightarrow c} \left(x\right)\right)}^{2} + 8 {\lim}_{x \rightarrow c} \left(x\right) - {\lim}_{x \rightarrow c} \left(15\right)}$

$= \sqrt{- {\left(c\right)}^{2} + 8 \left(c\right) - \left(15\right)}$

$= g \left(c\right)$

Use the one-sided versions of the limit properties at the endpoints.

For $c = 3$, replace all limits of the form ${\lim}_{x \rightarrow c}$ with ${\lim}_{x \rightarrow {3}^{+}}$

For $c = 5$, replace all limits of the form ${\lim}_{x \rightarrow c}$ with ${\lim}_{x \rightarrow {5}^{-}}$