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

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

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

By the power rule: 

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

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

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 $\displaystyle sec(2\pi)=1$. 

$\displaystyle y_1=0*0.01+1=1$

1 is our final answer. 

To do this, we must determine the slope of the function at $\displaystyle x=0$, which we will call $\displaystyle m$, and the initial value of the function at $\displaystyle x=0$, which we will call $\displaystyle b$, and since $\displaystyle x=0.02$ is only $\displaystyle .02$ away from $\displaystyle x=0$, our linear approximation will look like:

$\displaystyle y=m(.02)+b$

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

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

At $\displaystyle x=0$, our value for $\displaystyle m$ is $\displaystyle -3$

To determine $\displaystyle b$, we need to determine the value of the original equation at $\displaystyle x=0$

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

At $\displaystyle x=0$, our value for b is $\displaystyle 4$ 

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

First recall that

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

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

Recall that derivatives of exponential functions involving $\displaystyle e$ are given as:

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

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

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

At $\displaystyle x=-2$,

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

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

$\displaystyle y=3e^{-x}$

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

Therefore,

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

At $\displaystyle x=ln(e^3)=3$, 

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

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