Calculus 1 : How to find the meaning of functions

Study concepts, example questions & explanations for Calculus 1

varsity tutors app store varsity tutors android store

Example Questions

← Previous 1 3 4 5 6 7 8

Example Question #1 : How To Find The Meaning Of Functions

Take the limit

\lim_{x \to -2} \frac{x+\sqrt{x+6}}{x+2}

Possible Answers:

Correct answer:

Explanation:

First, multiply the numerator and denominator by x-\sqrt{x+6} and it turns into

\lim_{x \to -2} \frac{(x+\sqrt{x+6})}{(x+2)}\frac{(x-\sqrt{x+6})}{(x-\sqrt{x+6})}

\lim_{x \to -2} \frac{x^2-x-6}{(x+2)(x-\sqrt{x+6})}

Factor the numerator and then cancel out the 'x+2'

\lim_{x \to -2} \frac{(x+2)(x-3)}{(x+2)(x-\sqrt{x+6})}

\lim_{x \to -2} \frac{(x-3)}{(x-\sqrt{x+6})}

After taking the limit, the answer is

Example Question #2 : How To Find The Meaning Of Functions

If this limit is true, then what is the value of 'a'?

\lim_{x \to 3} \frac{x^4-a}{x^2-\sqrt{a}}=18

Possible Answers:

Correct answer:

Explanation:

Factor the numerator

\lim_{x \to 3} \frac{(x^2-\sqrt{a})(x^2+\sqrt{a})}{x^2-\sqrt{a}}=18

Cancel the x^2-\sqrt{a}, plug in the limit and then solve for 'a'

\lim_{x \to 3} x^2+\sqrt{a}=18

3^2+\sqrt{a}=18

a=81

Example Question #2 : How To Find The Meaning Of Functions

We have a line described as y=-\frac{1}{2}x+2. Find the minimum distance between the origin and a point on that line.

Possible Answers:

d=\frac{\sqrt{5}}{4}

d=\frac{4}{\sqrt{5}}

d=\sqrt{\frac{4}{5}}

1

\frac{16}{5}

Correct answer:

d=\frac{4}{\sqrt{5}}

Explanation:

We have the origin  and a point  located on the line.  That point represents the minimum distance to the orgin.  Apply the distance formula to these two points,

d=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}

Plug in the line equation, take the derivative, set it equal to zero, and solve for x.

d=\sqrt{x^2+(-\frac{1}{2}x+2)^2}=\sqrt{\frac{5}{4}x^2-2x+4}{d}'=\frac{\frac{5}{2}x-2}{2\sqrt{\frac{5}{4}x^2-2x+4}}=0

x=\frac{4}{5}

Use this  value to find

y=-\frac{1}{2}\left ( \frac{4}{5} \right )+2=\frac{8}{5}

So we have the point (\frac{4}{5},\frac{8}{5}), which is closest to the origin. We can now find its distance from that origin.

d=\sqrt{\left (\frac{4}{5} \right )^2+\left ( \frac{8}{5} \right )^2}

d=\sqrt{\frac{16}{5}}=\frac{4}{\sqrt{5}}

Example Question #3 : Meaning Of Functions

We have the following,

\int_{-5}^{5}\frac{x^2+(6+c)x+6c}{x+6}dx=-60

What is c?

Possible Answers:

Correct answer:

Explanation:

First, factor the numerator of the integrand.

\int_{-5}^{5}\frac{(x+6)(x+c)}{x+6}dx

Cancel out

\int_{-5}^{5}\frac{(x+6)(x+c)}{x+6}dx=-60

\int_{-5}^{5}(x+c)dx=-60

Perform the integral and then solve for

\frac{1}{2}(5)^2+5c-\left ( \frac{1}{2}(-5)^2+(-5)c \right )=-60

10c=-60

c=-6

Example Question #5 : How To Find The Meaning Of Functions

If , find 

Possible Answers:

Correct answer:

Explanation:

Taking the derivative of an integral yields the original function, but because we have a different variable in the integration limits, the variable switches

Example Question #1 : How To Find The Meaning Of Functions

Evaluate 

Possible Answers:

Correct answer:

Explanation:

using integration identities:  

Example Question #5 : Meaning Of Functions

Evaluate

Possible Answers:

Correct answer:

Explanation:

         evaluate at

Example Question #7 : How To Find The Meaning Of Functions

Evaluate 

Possible Answers:

Correct answer:

Explanation:

Intergation by substitution

new endpoints:

New Equation:

 at

Example Question #2 : How To Find The Meaning Of Functions

What is  ?

Possible Answers:

undefined

0

1

-1

Correct answer:

0

Explanation:

The relationship between  and x is an exponential relationship;  is going to  exponentially faster than  is going to  . One way to prove this is to write  and use L'Hôpital's rule:

 

 

Example Question #9 : How To Find The Meaning Of Functions

Where is  discontinuous? Are those discontinuities removable?

Possible Answers:

Essential discontinuities at  and .

Removable discontinuity at , essential discontinuity at .

Removable discontinuity at , essential discontinuity at

Removable discontinuities at  and

Removable discontinuities at and .

Correct answer:

Removable discontinuity at , essential discontinuity at

Explanation:

The rational function   has a denominator with two roots,  and . These are discontinuities.

 

Factoring both top and bottom and canceling a term  tells us that this function is equal to 

 

except at . This point is a removable discontinuity.  is therefore an essential discontinuity where the ratio goes to .

← Previous 1 3 4 5 6 7 8
Learning Tools by Varsity Tutors

Incompatible Browser

Please upgrade or download one of the following browsers to use Instant Tutoring: