Math 181 Miniproject 4: Linear Approximation and Calculus.md
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Math 181 Miniproject 4: Linear Approximation and Calculus
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**Overview:** In this miniproject you will put the idea of the *local linearization* of a function to build linear approximations to complex functions and then make *interpolations* and *extrapolations* using them.

**Prerequisites:** Sections 1.8 in *Active Calculus*, which focuses on this topic. **Completion of Miniprojects 1 and 2 is recommended before doing this miniproject**.

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1\. A potato is placed in an oven, and the potato's temperature $F$ (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. The time $t$ is measured in minutes.
| $t$ | 0 | 15 | 30 | 45 | 60 | 75 | 90 |
|----- |---- |------- |----- |----- |------- |------- |------- |
| $F$ | 70 | 180.5 | 251 | 296 | 324.5 | 342.8 | 354.5 |

(a) Use a central difference to estimate $F'(75)$. Use this estimate as needed in subsequent questions in this problem.
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(a) $$f'(75)= \frac{f(90)-f(60)}{90-60}$$
$$f'(75)=\frac{354.5-324.5}{30}$$
$$f'(75)=\frac{30}{30}$$
$$f'(75)=1\frac{ degF}{min}$$

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(b) Find the local linearization $y = L(t)$ to the function $y = F(t)$ at the point where $a = 75$.
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(b)$$L(x)= f(a)+f'(a)(x-a)$$
$$L(x)= f(75)+f'(75)(x-75)$$
$$L(x)=342.8+1(x-75)$$

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(c\) Determine an estimate for $F(72)$ by employing the local linearization. Terminology: This estimate is called an *interpolation* because we are estimating a value that lies within a data set, between two known data points.
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(c\)$$L(x)=342.8+1(x-75)$$
$$L(72)=342.8+1(72-75)$$
$$L(x)=342.8+1(-3)$$
$$L(x)=342.8-3$$
$$L(x)=339.8 deg F$$

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(d) Do you think your estimate in (c) is too large, too small, or exactly right? Why?
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(d)Since the value I calculated isn't way above f(75) or below f(60) it should be accurate enough. The number is a little more than f(75) but that's also because f(72) should have a close estimate.

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(e) Use your local linearization to estimate $F(100)$. Terminology: This estimate is called an *extrapolation* because we are estimating a value that lies outside the range of values of a data set.
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(e)$$L(x)=342.8+1(x-75)$$
$$L(x)=342.8+1(100-75)$$
$$L(x)=342.8+1(25)$$
$$L(x)=342.8+25$$
$$L(x)=370.8 deg F$$

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(f) Do you think your estimate in (e) is too large, too small, or exactly right? Why?
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(f) Considering it's supposed to be the value of f(100) I think that the estimate is too large. L(x) is the equation fro the tangent line of value f(75), which means that any number that's an outlier won't be accurate.Also, looking at the graph we can see that it's concave down while the tangent line is linear and f(100) isn't either of those; it doesn't fit into the function.

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(g) Plot both $F$ and $L$ and comment on how or when the line $L(t)$ is a good approximation of $F(t)$.
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(g) When f(x) is close to our reference point that's the only time line L(x) can be useful. Any other time won't woke because it won't match up to the concave down function since the tangent line is linear and going up.

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