Subjects
PRICING
COURSES
SIGN IN
Start Free Trial
Hunter F.
Engineer by day, Math tutor by night
Tutor Satisfaction Guarantee
Pre-Calculus
TutorMe
Question:

Knowing that the domain of a given function is all real numbers for which the function is defined, state the domain for f(x) = 2/(x-5).

Hunter F.
Answer:

D: (-infinity, 5) U (5, infinity)

Calculus
TutorMe
Question:

Find the linearization L(x) of f(x) = x^(2) + 2x at x = 1.

Hunter F.
Answer:

To start, find f(1). f(1) = 1^(2) + 2(1) = 3. Next, find f'(x). f'(x) = 2x + 2. Next, find f'(1). f'(1) = 2(1) + 2 = 4. Knowing that L(x) = f(a) + f'(a)(x-a), where x=a. Since x = 1, f(1) = 3, f'(x) = 2x + 2, and f'(1) = 4 L(x) = 3 + 4(x-1) = 4x - 1.

Algebra
TutorMe
Question:

Explain how the graphs of f(x) = 3^x and g(x) = 3^(-x) are similar and dissimilar to one another.

Hunter F.
Answer:

These graphs are similar due to their growth factor, which is 3. They also share the same y-intercept at (0,1). Both of these graphs have a horizontal asymptote at x=0. These graphs are dissimilar because of where x approaches. For the graph of f(x)=3^x, x approaches negative infinity, where the graph of g(x)=3^(-x), x approaches positive infinity. Additionally, f(x)=3^x is an example of exponential growth and g(x)=3^(-x) is exponentially decaying.

Send a message explaining your
needs and Hunter will reply soon.
Contact Hunter
Ready now? Request a lesson.
Start Session
FAQs
What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Session" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.