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Question S

f be the function defined by

f(~+1 for 0 3

f(~ 1 5 - ~ for 3 5,

Is f continuous at ~ - 3? ~xplain why or why not.

Find the average value of f(~) on the closed interval O 5.

Suppose the function g is defined by ~

| k~~ + 1 for 0 3

9(~ . mx + 2 for 3 5,

where k and m are constants. If g is ~ifferentiable at ~ - 3, what

Let

(a)

(b)

(c)

are the values of and m

(a) f is continuous at x - 3 because

lim f(x) - lim f(~) - 2-

x~3- x~3+

Therefore, ~~ f(~) - 2 - f(3),

1 answers "yes" and equates the

values of the left- and right-hand

2

limits

1 explanation involving limits

(b)

antiderivative of ~~

antiderivative of 5 - ~

evaluation and answer

2k - 3m + 2

k ~

-- 'III

4

values for k an

(c)

2.

Since g is continuous at ~ - 3, 2k - 3m +

k for O ~ 3

g'(x) 2 x +~

m for 3 ~ 5

xlim~3_ g'(~ 4-k and ~1~3+m 9~ - m

Since these two limits exist and g is

differentiable at x - 3 the two limits are

equal, Thus k - m.

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