Applications of Derivatives

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For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

$f(x)= \frac{x^{5}}{3}+2$

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=x+3\sin(x)$

$f(x)=\ln|x+3|$

Explanation

Recall the Maclaurin series formula:

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

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

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 $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

$f(x)= \frac{x^{5}}{3}+2$

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=x+3\sin(x)$

$f(x)=\ln|x+3|$

Explanation

Recall the Maclaurin series formula:

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

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

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

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\left ( -5, 5\right )$

$\left ( \frac{-1}{5}, \frac{1}{5} \right )$

$(-\infty, \infty)$

Explanation

Using the root test

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

and

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

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

This occurs when x=5.

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\left ( -5, 5\right )$

$\left ( \frac{-1}{5}, \frac{1}{5} \right )$

$(-\infty, \infty)$

Explanation

Using the root test

and

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

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

This occurs when x=5.

Write out the first four terms of the Taylor series about for the following function:

Explanation

The Taylor series about x=a of any function is given by the following:

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

So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

$f^{0}(x)=f(x)$

The derivatives were found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

$\frac{4(x-1)^0}{0!}+\frac{4(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}+\frac{0(x-1)^3}{3!}$

which when simplified is equal to

Write out the first four terms of the Taylor series about for the following function:

Explanation

The Taylor series about x=a of any function is given by the following:

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

So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

$f^{0}(x)=f(x)$

The derivatives were found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

$\frac{4(x-1)^0}{0!}+\frac{4(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}+\frac{0(x-1)^3}{3!}$

which when simplified is equal to

Find the minimum of the function:

$y=\frac{-3}{16}$

$y=\frac{3}{16}$

$y=\frac{-1}{8}$

$y=\frac{1}{8}$

Explanation

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

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

Solve for x.

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

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

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

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

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

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

Find the minimum of the function:

$y=\frac{-3}{16}$

$y=\frac{3}{16}$

$y=\frac{-1}{8}$

$y=\frac{1}{8}$

Explanation

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

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

Solve for x.

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

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

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

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

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

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

Write out the first three terms of the Taylor series for the following function about :

$0+(x-1)+\frac{(x-1)^2}{2}$

Explanation

The general formula for the Taylor series of a given function about x=a is

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

We were asked to find the first three terms, which correspond to n=0, 1, and 2. So first, we need to find the zeroth, first, and second derivative of the given function. The zeroth derivative is just the function itself.