CALCU-Question 11 ptsLet a R.Let f, g be two func

Solution by Steps

step 2

The second statement states that if

f

and

g

are not continuous at

a

, then

f+g

is not continuous at

a

. This is not necessarily true;

f

and

g

could have removable discontinuities that cancel each other out

step 3

The third statement suggests that if

f

has a REMOVABLE discontinuity at

a

and

g

has a REMOVABLE discontinuity at

a

, then

f+g

has a removable discontinuity at

0

. This is false; the discontinuities do not necessarily add up to create a removable discontinuity at

0

step 4

The fourth statement claims that if

f

has a NON-REMOVABLE discontinuity at

a

and

g

has a REMOVABLE discontinuity at

a

, then

f+g

has a non-removable discontinuity at

a

. This is true because the non-removable discontinuity of

f

dominates

step 5

The fifth statement states that if

\lim _{x \rightarrow a} f(x)=1

and

\lim _{x \rightarrow 1} g(x)=2

, then

\lim _{x \rightarrow a} g(f(x))=2

. This is true because

f(x)

approaches

1

, and thus

g(f(x))

approaches

g(1)=2

step 6

The sixth statement claims that if

\lim _{x \rightarrow a} f(x)

exists, then

\lim _{x \rightarrow a} e^{f(x)}

exists. This is true because the exponential function is continuous everywhere

[question number] Answer

A, D, E, F

Key Concept

Discontinuities and Limits

Explanation

Understanding the types of discontinuities (removable vs. non-removable) and their effects on limits and continuity is crucial in calculus.

Solution by Steps

step 1

We start with the limit we want to compute:

\lim_{x \rightarrow 0} \frac{f(4x)}{x}

step 2

We can rewrite this limit using the hint provided:

\lim_{x \rightarrow 0} \frac{f(4x)}{x} = \lim_{x \rightarrow 0} \frac{f(4x)}{4x} \cdot 4

step 3

From the problem, we know that

\lim_{x \rightarrow 0} \frac{f(x)}{x} = 3

. Therefore, we can substitute

4x

for

x

in this limit:

\lim_{x \rightarrow 0} \frac{f(4x)}{4x} = 3

step 4

Thus, we have

\lim_{x \rightarrow 0} \frac{f(4x)}{x} = 3 \cdot 4 = 12

Answer

12

Key Concept

Limits and continuity in functions

Explanation

The limit of a function as it approaches a point can often be simplified by substituting variables, allowing us to find the limit more easily. In this case, we used the known limit of

f(x)/x

to compute the desired limit.

Solution by Steps

step 1

To solve the equation

\frac{e^{x}+e^{-x}}{2}=2

on the interval

[-2,2]

, we first rewrite it as

e^{x} + e^{-x} = 4

step 2

The equation

e^{x} + e^{-x} = 4

can be transformed into

2\cosh(x) = 4

, leading to

\cosh(x) = 2

step 3

The solutions to

\cosh(x) = 2

are

x = \log(2 + \sqrt{3})

and

x = \log(2 - \sqrt{3})

. Since

\log(2 - \sqrt{3})

is not in the interval

[-2, 2]

, we only consider

x = \log(2 + \sqrt{3})

step 4

Therefore, the statement "There is no solution of

\frac{e^{x}+e^{-x}}{2}=2

on

[-2,2]

" is false

Answer

False

Solution by Steps

step 1

Given that

f

is a continuous function on

[0,1]

such that 0 < f(x) < 1 for all

x \in [0,1]

, we apply the Intermediate Value Theorem

step 2

Since f(0) > 0 and f(1) < 1 , there must exist some point

a \in [0,1]

such that

f(a) = a

by the Intermediate Value Theorem

step 3

Thus, the statement "We can conclude that there exists a point

a \in [0,1]

such that

f(a)=a

" is true

Answer

True

Solution by Steps

step 1

The statement claims that if a function

f

is continuous on

[a,b]

, then there exists a

c \in [a,b]

such that

f(c) = \frac{f(a) + f(b)}{2}

. This is a consequence of the Mean Value Theorem

step 2

Since

f

is continuous on the closed interval

[a,b]

, the function

f

must attain all values between

f(a)

and

f(b)

step 3

Therefore, the statement is true as there exists a

c

such that

f(c) = \frac{f(a) + f(b)}{2}

Answer

True

Solution by Steps

step 1

The function

f(x) = \frac{1+x^{2}}{x^{2}-4}

has vertical asymptotes at

x = -2

and

x = 2

where the denominator is zero

step 2

To find the maximum on the interval

[-3,3]

, we need to check the endpoints and any critical points within the interval

step 3

Since

f(x)

is undefined at

x = -2

and

x = 2

, we must evaluate the limits as

x

approaches these points. The function approaches

\infty

as

x \to -2^+

and

x \to 2^-

step 4

Therefore, the statement "The function

f(x)=\frac{1+x^{2}}{x^{2}-4}

has a maximum on

[-3,3]

" is false because it is undefined at the critical points

Answer

False

Key Concept

Understanding the properties of continuous functions and limits

Explanation

The answers are derived from applying theorems related to continuity and limits, confirming the truth or falsehood of each statement based on mathematical principles.

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