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)
$F'(75)=\frac{f(90)-f(60)}{90-60}$
$F'(x)=\frac{354.5-324.5}{90-60}$
$F'(75)=\frac{30}{30}$
$F'(75)=1\frac{°F}{min}$

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

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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)
I feel that the estimated value I concluded in the last part is fairly accurate because it isn't over the value of F(75) nor is it bellow F(60). It is just a bit bellow F(75), which I would expect since F(72) isn't considerably far away.

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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)
$L(x)=342.8+1(x-75)$
$L(100)=342.8+1(100-75)$
$L(100)=342.8+1(25)$
$L(100)=342.8+25$
$L(100)=370.8°F$

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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)
I believe that the my estimate in the previous problem is too big for the actual value of F(100). The reason why is because L(x) is just an equation for the tangent line at the value F(75) so anything too far from that, whether above or below, is not going to be accurate. Additionally, because the function seems to be concave down, the value F(100) would naturally be too far away from our tangent line, and therefore, this specific linear approximation would not work. In other words, the tangent line is a linear function while the actual funtion is a curve, so the estimated values of F(100) would be off either way.

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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)
I feel like the only time the line L(x) is a good approximation is when F(x) is close to our point of reference, anything too far away from it would be to inaccurate to estimate.

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