Enable contrast version

# Tutor profile: Dylan D.

Inactive
Dylan D.
Senior software developer in the financial sector with 10+ years of tutoring experience
Tutor Satisfaction Guarantee

## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Given the function $$f(x) = 2x^2 + 8x - 3$$, find the absolute maximum or minimum value and state whether it is a maximum or minimum.

Inactive
Dylan D.

The x-coordinate of the max. or min. value of a quadratic function $$f(x) = ax^2 + bx + c$$ is given by $$x = -\frac{b}{2a}$$. In our particular case, we see that $$a = 2$$ and $$b = 8$$. Therefore, $$x = -\frac{b}{2a} = -\frac{8}{2*2} = -2$$ Plugging this value back into $$f(x)$$, we find that $$f(-2) = 2(-2)^2 + 8(-2) - 3 = -11$$ We now know that $$-11$$ is the max. or min. value of $$f(x)$$ and that it occurs at $$x = -2$$. Next, we must determine whether it is a maximum or a minimum value. One good way of doing this is to graph some points near $$x = -2$$ on $$f(x)$$ and sketch the graph. It's important to understand the direction of the function on both sides of $$x = -2$$, as it will be a parabola extending infinitely in one vertical direction. To the left, we take the point $$x = -3$$, and find that $$f(-3) = 2(-3)^2 + 8(-3) - 3 = -9$$. To the right, we take the point $$x = 0$$, and find that $$f(0) = 2(0)^2 + 8(0) - 3 = -3.$$ In either case, we find that the value of $$f(x)$$ to the left or right of $$x = -2$$ is greater (less negative). Thus, based on our understanding of this particular function, and the general shape of quadratic functions, $$-11$$ is the minimum value of $$f(x)$$.

### Subject:Geometry

TutorMe
Question:

A square of side length $$s = 5$$ is inscribed in a circle. Determine the circumference C of the circle.

Inactive
Dylan D.

Our goal is to solve for the circumference of the circle, which is given by $$C = 2 \pi r$$. However, we can solve for the circumference only once we know the radius of the circle. We are told that the square is inscribed in the circle, which means that each corner of the square is touching the circle. We can see from visual inspection that this means that the diagonal of the square is the same as the diameter of the circle. A square can be divided into two separate right isosceles triangles. For either one, the hypotenuse, which is the diagonal of the square, is given by $$h = s * \sqrt{2}$$, where $$s$$ is the side length of the square. Since we are given that $$s = 5$$, we can compute that $$h = 5 \sqrt{2}$$. Since the diagonal of the inscribed square has length $$h = 5 \sqrt{2}$$, the diameter of the circle $$d = 5 \sqrt{2}$$, as well. Our equation for circumference is written in terms of the radius, so we divide the diameter in half to obtain $$r = \frac{5}{2} \sqrt{2}$$. Finally, we solve for the circumference: $$C = 2 \pi r = 2 \pi(\frac{5}{2} \sqrt{2}) = 5 \pi \sqrt{2}$$

### Subject:Calculus

TutorMe
Question:

Given the derivative $$f'(x) = 8x + 4$$ and the fact that $$f(0) = 6$$, determine the equation $$f(x)$$.

Inactive
Dylan D.

We are given the first derivative, $$f'(x)$$, and are tasked with determining the original function $$f(x)$$. From the fundamental theorem of calculus, we know that we can use integration to determine $$f(x)$$ up to a constant: $$\int f'(x)dx = f(x) + C$$, for some yet unknown constant C Thus: $$\int f'(x)dx = \int (8x + 4)dx = 4x^2 + 4x + C$$ Now, we need more information to determine the constant C. Specifically, we need to know a pair of values that correspond to $$f(x)$$. In fact, we are provided this information in the question. Namely, we know that $$f(0) = 6$$. Therefore: $$f(0) = 6 \implies 4*(0)^2 + 4*0 + C = 6 \implies C = 6$$ Now that we know the value of the constant C, we can replace C in our earlier equation. We have successfully determined that $$f(x) = 4x^2 + 4x + 6$$.

## Contact tutor

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

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