* SOLUTION 1 :* Function *f* is defined at *x*=1 since

i.) *f*(1) = 2 .

The limit

= 3 (1) - 5

= -2 ,

i.e.,

ii.) .

But

iii.) ,

so condition iii.) is not satisfied and function *f* is NOT continuous at *x*=1 .

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* SOLUTION 2 :* Function *f* is defined at *x*=-2 since

i.)
*f*(-2) = (-2)^{2} + 2(-2) = 4-4 = 0 .

The left-hand limit

= (-2)^{2} + 2(-2)

= 4 - 4

= 0 .

The right-hand limit

= (-2)^{3} - 6(-2)

= -8 + 12

= 4 .

Since the left- and right-hand limits are not equal, ,

ii.) does not exist,

and condition ii.) is not satisfied. Thus, function *f* is NOT continuous at *x*=-2 .

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* SOLUTION 3 :* Function *f* is defined at *x*=0 since

i.) *f*(0) = 2 .

The left-hand limit

= 2 .

The right-hand limit

= 2 .

Thus, exists with

ii.) .

Since

iii.) ,

all three conditions are satisfied, and *f* is continuous at *x*=0 .

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* SOLUTION 4 :* Function *h* is not defined at *x*=-1 since it leads to division by zero. Thus,

i.) *h*(-1)

does not exist, condition i.) is violated, and function *h* is NOT continuous at *x* = -1 .

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* SOLUTION 5 :* First, check for continuity at *x*=3 . Function *f* is defined at *x*=3 since

i.) .

The limit

(Circumvent this indeterminate form by factoring the numerator and the denominator.)

(Recall that
*A*^{2} - *B*^{2} = (*A*-*B*)(*A*+*B*) and
*A*^{3} - *B*^{3} = (*A*-*B*)(*A*^{2}+*AB*+*B*^{2} ) . )

(Divide out a factor of (*x*-3) . )

=

,

i.e.,

ii.) .

Since,

iii.) ,

all three conditions are satisfied, and *f* is continuous at *x*=3 . Now, check for continuity at *x*=-3 . Function *f* is not defined at *x* = -3 because of division by zero. Thus,

i.) *f*(-3)

does not exist, condition i.) is violated, and *f* is NOT continuous at *x*=-3 .

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* SOLUTION 6 :* Functions
*y* = *x*^{2} + 3*x* + 5 and
*y* = *x*^{2} + 3*x* - 4 are continuous for all values of *x* since both are polynomials. Thus, the quotient of these two functions,
,
is continuous for all values of *x* where the denominator,
*y* = *x*^{2} + 3*x* - 4 = (*x*-1)(*x*+4) , does NOT equal zero. Since
(*x*-1)(*x*+4) = 0 for *x*=1 and *x*=-4 , function *f* is continuous for all values of *x* EXCEPT *x*=1 and *x*=-4 .

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* SOLUTION 7 :* First describe function *g* using functional composition. Let
*f*(*x*) = *x*^{1/3} ,
, and
*k*(*x*) = *x*^{20} + 5 . Function *k* is continuous for all values of *x* since it is a polynomial, and functions *f* and *h* are well-known to be continuous for all values of *x* . Thus, the functional compositions

and

are continuous for all values of *x* . Since

,

function *g* is continuous for all values of *x* .

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* SOLUTION 8 :* First describe function *f* using functional composition. Let
*g*(*x*) = *x*^{2} - 2*x* and
. Function *g* is continuous for all values of *x* since it is a polynomial, and function *h* is well-known to be continuous for
. Since
*g*(*x*) = *x*^{2} - 2*x* = *x*(*x*-2) , it follows easily that
for
and
. Thus, the functional composition

is continuous for and . Since

,

function *f* is continuous for
and
.

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* SOLUTION 9 :* First describe function *f* using functional composition.
Let
and
. Since *g* is the quotient of polynomials
*y* = *x*-1 and *y* = *x*+2 , function *g* is continuous for all values of *x* EXCEPT where *x*+2 = 0 , i.e., EXCEPT for *x* = -2 . Function *h* is well-known to be continuous for *x* > 0 . Since
, it follows easily that *g*(*x*) > 0 for *x* < -2 and *x* > 1 . Thus, the functional composition

is continuous for *x* < -2 and *x* > 1 . Since

,

function *f* is continuous for *x* < -2 and *x* > 1 .

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* SOLUTION 10 :* First describe function *f* using functional composition.
Let
and
*h*(*x*) = *e*^{ x } , both of which are well-known to be continuous for all values of *x* . Thus, the numerator
is continuous (the functional composition of continuous functions) for all values of *x* . Now consider the denominator
. Let *g*(*x*) = 4 ,
*h*(*x*) = *x*^{2} - 9 , and
. Functions *g* and *h* are continuous for all values of *x* since both are polynomials, and it is well-known that function *k* is continuous for
. Since
*h*(*x*) = *x*^{2} - 9 = (*x*-3)(*x*+3) = 0 when *x*=3 or *x*=-3 , it follows easily that
for
and
, so that
is continuous (the functional composition of continuous functions) for
and
. Thus, the denominator
is continuous (the difference of continuous functions) for
and
. There is one other important consideration. We must insure that the DENOMINATOR IS NEVER ZERO. If

then

.

Squaring both sides, we get

16 = *x*^{2} - 9

so that

*x*^{2} = 25

when

*x* = 5 or *x* = -5 .

Thus, the denominator is zero if *x* = 5 or *x* = -5 . Summarizing, the quotient of these continuous functions,
, is continuous for
and
, but NOT for *x* = 5 and *x* = -5 .

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* SOLUTION 11 :* Consider separately the three component functions which determine *f* . Function
is continuous for *x* > 1 since it is the quotient of continuous functions and the denominator is never zero. Function *y* = 5 -3*x* is continuous for
since it is a polynomial. Function
is continuous for *x* < -2 since it is the quotient of continuous functions and the denominator is never zero. Now check for continuity of *f* where the three components are joined together, i.e., check for continuity at *x*=1 and *x*=-2 . For *x* = 1 function *f* is defined since

i.)
*f*(1) = 5 - 3(1) = 2 .

The right-hand limit

=

(Circumvent this indeterminate form one of two ways. Either factor the numerator as the difference of squares, or multiply by the conjugate of the denominator over itself.)

= 2 .

The left-hand limit

=

= 5 - 3(1)

= 2 .

Thus,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and function *f* is continuous at *x*=1 . Now check for continuity at
*x*=-2 . Function *f* is defined at *x*=-2 since

i.)
*f*(-2) = 5 - 3(-2) = 11 .

The right-hand limit

=

= 5 - 3( -2)

= 11 .

The left-hand limit

=

= -1 .

Since the left- and right-hand limits are different,

ii.) does NOT exist,

condition ii.) is violated, and function *f* is NOT continuous at *x*=-2 . Summarizing, function *f* is continuous for all values of *x* EXCEPT *x*=-2 .

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* SOLUTION 12 :* First, consider separately the two components which determine function *f* . Function
*y* = *A*^{2} *x* - *A* is continuous for
for any value of *A* since it is a polynomial. Function *y* = 4 is continuous for *x* < 3 since it is a polynomial. Now determine *A* so that function *f* is continuous at *x*=3 . Function *f* must be defined at *x*=3 , so

i.)
*f*(3)= *A*^{2} (3) - *A* = 3 *A*^{2} - *A* .

The right-hand limit

=

= *A*^{2} (3) - *A*

= 3 *A*^{2} - *A* .

The left-hand limit

=

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

so that

3 *A*^{2} - *A* - 4 = 0 .

Factoring, we get

(3*A* - 4)(*A* + 1) = 0

for

or *A* = -1 .

For either choice of *A* ,

iii.) ,

all three conditions are satisfied, and *f* is continuous at *x*=3 . Therefore, function *f* is continuous for all values of *x* if
or *A* = -1 .

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* SOLUTION 13 :* First, consider separately the three components which determine function *f* . Function
*y* = *Ax* - *B* is continuous for
for any values of *A* and *B* since it is a polynomial. Function
*y* = 2*x*^{2} + 3*Ax* + *B* is continuous for
for any values of *A* and *B* since it is a polynomial. Function *y* = 4 is continuous for *x* > 1 since it is a polynomial. Now determine *A* and *B* so that function *f* is continuous at *x*=-1 and *x*=1 . First, consider continuity at *x*=-1 . Function *f* must be defined at *x*=-1 , so

i.)
*f*(-1)= *A*(-1) - *B* = - *A* - *B* .

The left-hand limit

=

= *A* (-1) - *B*

= - *A* - *B* .

The right-hand limit

=

= 2(-1)^{2} + 3*A*(-1) + *B*

= 2 - 3*A* + *B* .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

so that

2*A* - 2*B* = 2 ,

or

(Equation 1)

*A* - *B* = 1 .

Now consider continuity at *x*=1 . Function *f* must be defined at *x*=1 , so

i.)
*f*(1)= 2(1)^{2} + 3*A*(1) + *B* = 2 + 3*A* + *B* .

The left-hand limit

=

= 2(1)^{2} + 3*A*(1) + *B*

= 2 + 3*A* + *B* .

The right-hand limit

=

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

or

(Equation 2)

3*A* + *B* = 2 .

Now solve Equations 1 and 2 simultaneously. Thus,

*A* - *B* = 1 and
3*A* + *B* = 2

are equivalent to

*A* = *B* + 1 and
3*A* + *B* = 2 .

Use the first equation to substitute into the second, getting

3 (*B* + 1 ) + *B* = 2 ,

3 *B* + 3 + *B* = 2 ,

and

4 *B* = -1 .

Thus,

and

.

For this choice of *A* and *B* it can easily be shown that

iii.)

and

iii.) ,

so that all three conditions are satisfied at both *x*=1 and *x*=-1 , and function *f* is continuous at both *x*=1 and *x*=-1 . Therefore, function *f* is continuous for all values of *x* if
and
.

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* SOLUTION 14 :* First describe *f* using functional composition. Let
*g*(*x*) = -1/*x*^{2} and
*h*(*x*) = *e*^{x} . Function *h* is well-known to be continuous for all values of *x* . Function *g* is the quotient of functions continuous for all values of *x* , and is therefore continuous for all values of *x* except *x*=0 , that *x* which makes the denominator zero. Thus, for all values of *x* except *x*=0 ,

*f*(*x*) = *h* ( *g*(*x*) ) = *e*^{ g(x) } = *e*^{ -1/x2 }

is a continuous function (the functional composition of continuous functions). Now check for continuity of *f* at *x*=0 . Function *f* is defined at *x*=0 since

i.) *f*(0) = 0 .

The limit

(The numerator approaches -1 and the denominator is a positive number approaching zero.)

,

so that

= 0 ,

i.e.,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and *f* is continuous at *x*=0 . Thus, *f* is continuous for all values of *x* .

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* SOLUTION 15 :* First show that *f* is continuous for all values of *x* . Describe *f* using functional composition. Let
,
, and
*k*(*x*) = *x*^{2} . Function *h* is well-known to be continuous for all values of *x* . Function *k* is a polynomial and is therefore continuous for all values of *x* . Function *g* is the quotient of functions continuous for all values of *x* , and is therefore continuous for all values of *x* except *x*=0 , that *x* which makes the denominator zero. Thus, for all values of *x* except *x*=0 ,

is a continuous function (the product and functional composition of continuous functions). Now check for continuity of *f* at *x*=0 . Function *f* is defined at *x*=0 since

i.) *f*(0) = 0 .

The limit
does not exist since the values of
oscillate between -1 and +1 as *x* approaches zero.
However, for

so that

.

Since

,

it follows from the Squeeze Principle that

ii.) .

Since

iii.) ,

all three conditions are satisfied, and *f* is continuous at *x*=0 . Thus, *f* is continuous for all values of *x* .
Now show that *f* is differentiable for all values of *x* . For
we can differentiate *f* using the product rule and the chain rule. That is, for
the derivative of *f* is

.

Use the limit definition of the derivative to differentiate *f* at *x*=0 . Then

.

Use the Squeeze Principle to evaluate this limit. For

.

If , then

.

If , then

.

In either case,

,

and it follows from the Squeeze Principle that

.

Thus, *f* is differentiable for all values of *x* . Check to see if *f*' is continuous at *x*=0 . The function *f*' is defined at *x*=0 since

i.) *f*'(0) = 0 .

However,

ii.)

does not exist since the values of
oscillate between -1 and +1 as *x* approaches zero. Thus, condition ii.) is violated, and the derivative , *f*' , is not continuous at *x*=0 .

NOTE : The continuity of function *f* for all values of *x* also follows from the fact that *f* is differentiable for all values of *x* .

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