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)
$f'(75)=(f(90)-f(60))/(90-60)$
$f'(75)=(354.5-324.5)/30=1 degF^o/min$

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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)
$y=L(x)=f(75)+f'(75)(x-75)$
$L(x)=342.8+1(x-75)$

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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\)
$F(72)=L(72)=342.8+1(72-75)$
$L(72)=342.8-3=339.8$

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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)
I feel it is pretty accurate. Maybe not exactly right but pretty close, because it falls under f(75) and above f(60).

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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)
$f(100)=L(100)=342.8+1(100-75$)
$L(100)=342.8+25=367.8$
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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)
I feel like it is esactly correct because it is not to high away from f(90).

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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)


L(t) is a close estimate of F(t) (black line). The L(t) (orange line) values were not 100% on the function; however, they were close enough to give us a close representitive to what we should expect for our f(72) and f(100) values. So, instead of saying L(t)=f(t) we can say that L(t) is just about, or a rough estimate of f(t).
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