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]?

There is no one sentence answer.

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

For ##c## in ##(3,5)##, we need##lim_(xrarrc) g(x) = g(c)## and we also need one-sided continuity at the endpoints: we need: ## lim_(xrarr3^+) g(x) = g(3)## and ##lim_(xrarr5^-) g(x) = g(5)##

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

##lim_(xrarrc) g(x) = lim_(xrarrc) sqrt(-x^2+8x-15)##

##= sqrt(lim_(xrarrc)(-x^2+8x-15))##

##= sqrt(lim_(xrarrc)(-x^2)+lim_(xrarrc)(8x)-lim_(xrarrc)(15))##

##= sqrt(-lim_(xrarrc)(x^2)+8lim_(xrarrc)(x)-lim_(xrarrc)(15))##

##= sqrt(-(lim_(xrarrc)(x))^2+8lim_(xrarrc)(x)-lim_(xrarrc)(15))##

##= sqrt(-(c)^2+8(c)-(15))##

##= g(c)##

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

For ##c=3##, replace all limits of the form ##lim_(xrarrc)## with ##lim_(xrarr3^+)##

For ##c=5##, replace all limits of the form ##lim_(xrarrc)## with ##lim_(xrarr5^-)##