Enable contrast version

# Tutor profile: Chantelle H.

Inactive
Chantelle H.
Friendly Math Tutor with Master's Degree in Mathematics
Tutor Satisfaction Guarantee

## Questions

### Subject:Linear Algebra

TutorMe
Question:

Let $$A,B,C,X$$ be invertible square matrices of the same size. Solve for $$X$$ in terms of $$A$$, $$B$$, and $$C$$ in the equation $$AXB+BC=C$$.

Inactive
Chantelle H.

Start with $$AXB+BC=C$$. Multiply both sides by $$A^{-1}$$ on the left to get $$A^{-1}AXB+A^{-1}BC=A^{-1}C$$ Remember: It is important to choose to multiply $$A^{-1}$$ on the right or the left as $$MN\not= NM$$ for matrices (or in fancy math language, matrix multiplication is not commutative). Simplify using $$A^{-1}A=I$$ to get $$XB+A^{-1}BC=A^{-1}C$$ Multiply by $$B^{-1}$$ on the left to get $$XBB^{-1}+A^{-1}BCB^{-1}=A^{-1}CB^{-1}$$ Simplify using $$BB^{-1}=I$$ to get $$X+A^{-1}BCB^{-1}=A^{-1}CB^{-1}$$ Subtract $$A^{-1}BCB^{-1}$$f from both sides to get $$X=A^{-1}CB^{-1}-A^{-1}BCB^{-1}$$ Bonus Step: This is not necessary but if you feel like being fancy you get $$X=A^{-1}(I-B)(CB^{-1})$$ (factor through to see why this is true).

### Subject:Pre-Calculus

TutorMe
Question:

Let $$f(x)=x-3$$ and $$g(x)=sin(x)$$. Compute $$(f\circ g)(x)$$ and $$(f\circ g)(0)$$.

Inactive
Chantelle H.

Step 1: What does $$(f\circ g)(x)$$ mean? $$(f\circ g)(x)$$ is the function composition of $$f$$ and $$g$$ evaluated at the point $$x$$. It is pronounced "$$f$$ of $$g$$ of $$x$$" and is equivalent to $$f(g(x))$$. That is first you evaluate $$g(x)$$ and then plug that number into the function for $$f$$. Step 2: Brackets I find it helpful to start by putting brackets around all the $$x$$'s in $$f(x)$$. This does not change the meaning, but ensures we don't make any errors when we sub in $$g(x)$$. Here $$f(x)=(x)-3$$ Step 3: Solve $$(f\circ g)(x)=f(g(x))$$. To do this substitute all your $$(x)'s$$ with $$(g(x))'s$$. Therefore we have $$(f\circ g)(x)=f(g(x))=(g(x))-3=(sin(x)-3)$$ Step 4: Solve $$(f\circ g)(0)$$. There are two ways to do this. a) Plug 0 into our equation for $$(f\circ g)(0)=sin(0)-3=0-3=-3$$ b) First solve $$g(0)=sin(0)=0$$ and plug the result into $$f(g(0))=f(0)=0-3=-3$$.

### Subject:Calculus

TutorMe
Question:

Find the tangent line to the function $$f(x)=x^2+1$$ at the point $$x=2$$.

Inactive
Chantelle H.

Step 1: What is a tangent line? A tangent to the function $$f(x)$$ at the point $$x=a$$ is a line $$y=mx+b$$ that a) Passes through the point $$(a, f(a))$$. b) Has the same "slope" as $$f(x)$$ at $$a$$, or more precisely $$m=f'(a)$$ Step 2: What point does our tangent line have to pass through? The question asks us for the tangent line to $$f$$ at the point $$x=2$$. Therefore in this case $$a=2$$ and $$f(a)=f(2)=2^2+1=5$$. Therefore the correct point is $$(2,5)$$. Step 3: What is the slope of the tangent line? The slope of the tangent line is the same as $$f'(a)$$. We can use the rules $$(g+h)'=g'+h'$$ $$(x^n)'=nx^{n-1}$$ $$f'(c)=0$$ for any constant $$c\in \mathbb{R}$$ to determine $$f'(x)=2x$$. Therefore $$f'(a)=f'(2)=2a=2(2)=4$$. Step 4: The equation of the tangent line. We know that $$y=mx+b$$ is the tangent line with slope $$m=4$$ though the point $$(2,5)$$. We can substitute in $$(2,5)$$ to find the value of $$b$$. $$5=(4)(2)+b$$ Therefore $$b=5-8=-3$$ and $$y=4x-3$$ is the tangent line to $$f(x)=x^2+1$$ at $$x=2$$

## Contact tutor

Send a message explaining your
needs and Chantelle will reply soon.
Contact Chantelle

Start Lesson

## 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 Lesson" 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.
BEST IN CLASS SINCE 2015
TutorMe homepage