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Calculus 1 : How to find prediction models

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Example Question #1 : How To Find Prediction Models

Suppose you are a banker and set up a very unique function for your interest rate over time given by

$y=sec(\pi x)$

However, you find your computer incapable of calculating the interest rate at x=2.01 . Estimate the value of the interest rate at x=2.01 by using a linear approximation, using the slope of the function at x=2 .

Possible Answers:

Undefined

Correct answer:

Explanation:

To do a linear approximation, we're going to create a function

y_1=mz+b , that approximates our situation. In our case, m will be the slope of the function $sec(\pi x)$ at x=2 , while b will be the value of the function $sec(\pi x)$ at x=2 . The z will be distance from our starting position (x=2) to our end position (x=2.01) , which is 0.01 .

Firstly, we need to find the derivative of $sec(\pi x)$ with respect to x to determine slope.

By the power rule:

$\frac{\mathrm{d} }{\mathrm{d} x} (cos(\pi x))^{-1}= -(cos(\pi x))^{-2}*-\pi sin(\pi x)=(cos(\pi x))^{-2}*\pi sin(\pi x)$

The slope at x=2 will therefore be 0 since $sin(2\pi)=0$ .

Since this is the case, the approximate value of our interest rate will be identical to the value of the original function at x=2, which is $sec(2\pi)=1$ .

y_1=0*0.01+1=1

1 is our final answer.

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Example Question #2 : How To Find Prediction Models

Approximate the value at x=0.02 of the function $y=3e^{-x}+sec(x)$ ,with a linear approximation using the slope of the function at x=0 .

Possible Answers:

3.94

-3.94

Correct answer:

3.94

Explanation:

To do this, we must determine the slope of the function at x=0 , which we will call , and the initial value of the function at x=0 , which we will call , and since x=0.02 is only away from x=0 , our linear approximation will look like:

y=m(.02)+b

To determine slope, we take the derivative of the function with respect to x and find its value at x=0 , which in our case is:

$\frac{\mathrm{d} }{\mathrm{d} x}[3e^{-x}+sec(x)]= -3e^{-x}-sec(x)*tan(x).$

At x=0 , our value for is

To determine , we need to determine the value of the original equation at x=0

$y=3e^{-x}+sec(x)$

At x=0 , our value for b is

Since y=m(.02)+b , y=3.94

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Example Question #2 : How To Find Prediction Models

Determine the tangent line to $y=3e^{-x}$ at $x=ln(e^{-2})$ , and use the tangent line to approximate the value at x=ln(e^3) .

Possible Answers:

-6e^2

12e^2

e^2

-12e^2

Correct answer:

-12e^2

Explanation:

First recall that

$ln(e^{-2})=-2$

To find the tangent line of $y=3e^{-x}$ at x=-2 , we first determine the slope of $y=3e^{-x}$ . To do so, we must find its derivative.

Recall that derivatives of exponential functions involving are given as:

$\frac{\mathrm{d} }{\mathrm{d} x}Ae^{g(x)}= A*g'(x)e^{g(x)}$ , where is a constant and g(x) is any function of

In our case, A=3, g(x)=-x, g'(x)=-1 ,.

$\frac{d}{dx}3e^{-x}= -3e^{-x}$

At x=-2 ,

$\frac{\mathrm{d} y}{\mathrm{d} x}=-3e^{-(-2)}}= -3e^2=m$ , where is the slope of the tangent line.

To use point-slope form, we need to know the value of the original function at x=-2 ,

$y=3e^{-x}$

$y(-2)=3e^{-(-2)}=3e^2$

Therefore,

$(y_{appx}-3e^2)=-3e^2(x+2)$

At x=ln(e^3)=3 ,

$(y_{appx}-3e^2)=-3e^2(3+2)$

$y_{appx}=-15e^2+3e^2=-12e^2$

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Calculus 1 : How to find prediction models

Study concepts, example questions & explanations for Calculus 1

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Example Question #1 : How To Find Prediction Models

Example Question #2 : How To Find Prediction Models

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