To find the local max, you must find the first derivative, which is $\displaystyle y=-3x^2+3$.

Then. you need to set that equal to zero, so that you can find the critical points. The critical points are telling you where the slope is zero, and also clues you in to where the function is changing direction. When you set this derivative equal to zero and factor the function, you get $\displaystyle (x-1)(x+1)=0$, giving you two critical points at $\displaystyle x=1$ and $\displaystyle x=-1{}$.

Then, you set up a number line and test the regions in between those points. To the left of -1, pick a test value and plug it into the derivative. I chose -2 and got a negative value (you don't need the specific number, but rather, if it's negative or positive). In between -1 and 1, I chose 0 and got a possitive value. To the right of 1, I chose 2 and got a negative value. Then, I examine my number line to see where my function was going from positive to negative because that is what yields a maximum (think about a function going upwards and then changing direction downwards). That is happening at x=1.

When the derivative of a function is equal to zero, that means that the point is either a local maximum, local miniumum, or undefined. The derivative of $\displaystyle x^n$ is $\displaystyle nx^{n-1}$. The derivative of the given function is 

$\displaystyle f'(x)=3x^2-12x-36$

We must now set it equal to zero and factor.

$\displaystyle 3(x^2-4x-12)=0$

$\displaystyle x^2-4x-12=0$

$\displaystyle (x-6)(x+2)=0$

$\displaystyle x=6,-2$

Now we must plug in points to the left and right of the critical points to determine which is the local maximum. 

$\displaystyle f'(-3)=27$

$\displaystyle f'(0)=-36$

$\displaystyle f'(7)=27$

This means the local maximum is at $\displaystyle x=-2$ because the function is increasing at numbers less than -2 and decreasing at number between -2 and 6

The points where the derivative of a function are equal to 0 are called critical points. Critical points are either local maxs, local mins, or do not exist. The derivative of $\displaystyle x^n$ is $\displaystyle nx^{n-1}$. The derivative of the function is

$\displaystyle f'(x)=6x^2+30x-36$

Now we must set it equal to 0 and factor to solve.

$\displaystyle 6(x^2+5x-6)=0$

$\displaystyle x^2+5x-6=0$

$\displaystyle (x+6)(x-1)=0$

$\displaystyle x=-6,1$

We must now plug in points to the left and right of the critical points into the derivative function to figure out which is the local max.

$\displaystyle f'(-7)=48$

$\displaystyle f'(0)=-36$

$\displaystyle f'(2)=48$

This means that the function is increasing until it hits x=-6, then it decreases until x=1, then it begins increasing again.

This means that x=-6 is the local max.

At local maximums and minumims, the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative of the function.

The derivative of $\displaystyle x^n$ is $\displaystyle nx^{n-1}$. Thus the derivative of the function is 

$\displaystyle f'(x)=6x^2-18x+12$

To find maximums and minumums we set it equal to 0.

$\displaystyle 6x^2-18x+12=0$

$\displaystyle =6x^2-12x-6x+12$

$\displaystyle =6x(x-2)-6(x-2)$

$\displaystyle =(6x-6)(x-2)=0$

$\displaystyle x=1,2$

So the critical points are at x=1 and x=2. To figure out the maximum we must plug each into the original function.

$\displaystyle f(1)=2(1)^3-9(1)^2+12(1)-12=-7$

$\displaystyle f(2)=2(2)^3-9(2)^2+12(2)-12=-8$

So the local max is at x=1.

$\displaystyle f(x)=12x^2+3x$

To find minimum take the derivative of the function and set it equal to zero.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}=24x+3$

$\displaystyle 0=24x+3$

Solve for x.

$\displaystyle -3=24x$

$\displaystyle x=\frac{-1}{8}$

Plugging x back in the equation will allow us to find the y value that results in the minimum.

$\displaystyle f(\frac{-1}{8})=12(\frac{-1}{8})^2+3(\frac{-1}{8})$

$\displaystyle f(\frac{-1}{8})=\frac{3}{16}+(\frac{-3}{8})$

$\displaystyle f(\frac{-1}{8})=\frac{-3}{16}$

The graph has a minimum at y=-3/16

Recall the Maclaurin series formula:

$\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$\displaystyle =f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $\displaystyle f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is$\displaystyle 2+\frac{1}{3}x^5$.

Using the formula of a binomial series centered at 0:

 $\displaystyle (1+x)^k =\sum_{n=0}^{\infty } \binom{k}{n}x^n$,

where we replace $\displaystyle x$ with $\displaystyle \frac{x}{2}$ and $\displaystyle k=\frac{1}{3}$, we get:

 $\displaystyle \sum_{n=0}^{2}\binom{\frac{1}{3}}{n}\left(\frac{x}{2}\right)^n$ for the first 3 terms.

Then, we find the terms where,

 $\displaystyle T_2(x)=\binom{\frac{1}{3}}{0}(\frac{x}{2})^0 + \binom{\frac{1}{3}}{1}(\frac{x}{2})^1+\binom{\frac{1}{3}}{2}(\frac{x}{2})^2=1+\frac{x}{6}- \frac{x^2}{36}$

By the ratio test, the series $\displaystyle \sum_{n=0}^{\infty }\frac{5^n}{n!}$ converges absolutely: 

$\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{5^{n+1}}{(n+1)!} \cdot \frac{n!}{5^n} \right |=\lim_{n\rightarrow \infty }\left | \frac{5}{n+1} \right |=0< 1$

Using the root test

$\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$ 

and

$\displaystyle \lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$. T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

To get $\displaystyle f(x)$, we need to find the antiderivative of $\displaystyle f'(x)$ by integrating the third degree polynomial term by term.

We only want up to a third degree polynomial, so we can disregard the fourth order term:

$\displaystyle f(x) = c + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Since $\displaystyle f(2) = 7$, substitute $\displaystyle c=7$ for the final $\displaystyle f(x)$.

$\displaystyle f(x)=7 + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$