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]?

1 Answer
Jim H
Jun 15, 2015

There is no one sentence answer.

Explanation:

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^-)#

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