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Buddy G.

Tutoring and teaching math at College level for several years.

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Pre-Calculus

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Question:

What is the average rate of change for the function f(x)=x^2 on the interval [-1, 2]?

Buddy G.

Answer:

For any given function f(x) and an interval [a, b] in its domain the average rate of change for that function on that given interval is equal to [f(b) - f(a)] / (b-a). Thus, for this example we calculate the average rate of change to be [f(2) - f(-1)] / [2-(-1)] = [(2^2) - ((-1)^2)] / 3 =3/3 = 1 Note that this value gives the slope of the secant line going through the function's graph at the domain values of -1 and 2.

Trigonometry

TutorMe

Question:

Given a right triangle with an interior angle of 30, an adjacent leg length of 6 and an opposite leg length of 9; calculate the length of the hypotenuse without using the Pythagorean theorem.

Buddy G.

Answer:

First draw the picture and label all given values and let's label our unknown value for the length of the hypotenuse, x. We can solve this problem in two different ways with the information given. We can use either the sine or cosine trig functions. Let's use the sine function noting that sin(30)=1/2. [Note: make sure your calculator is set to degree mode] sin(30)=9/x giving us that x=9/sin(30)=9/(1/2)=18 So, our right triangle has a hypotenuse length of 18 units.

Calculus

TutorMe

Question:

Find the equation for the tangent line at x=3 for the function f(x) = 2x^2 +1.

Buddy G.

Answer:

The derivative of f(x) is f'(x) = 4x. Using our x-value of 3 gives us the slope of our tangent line by plugging into our formula for the derivative and gives us a point on the tangent line by plugging into our original function. So, f'(3)=4(3)=12 and f(3)=2(3)^2+1=19 which tells us the slope is 19 and a point on the line is (3, 19). Using point-slope formula we get the following: y-19=12(x-3) y-19=12x-36 y=12x-17 This is the equation we required.

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