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)(60,324.5) and (90, 354.5) are the points I used to find the central drifference which is 1.

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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) The local linerization at a=75 is 343.8898.
By plugging a=75 into the fucnction below F(x)= 343.944.
The formula I used what the tangent line of $F(x)=-304.952*0.970^x+374.944$ at a=75. Which is L(x)=0.9458615(x)+272.950.

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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\) From the local linerization formula used before, plugging in 72 equaled 341.05222
Plugging it into the function F(x)=340.919.
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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) My estimate is too large because by plugging 72 into the actual function of is 340.919.

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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)Using local linerization for the estimate F(100) is L(100)=367.5361 which is a fair amount off from F(100)=360.4427.
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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 think the estimate in part E for L(100) is larger than F(100).

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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 not a good representation of F(t) since the farther away from the point 75, the larger the disparity of the number L(x) gives versus what F(x) actually is.

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