Applications of Derivatives - GRE Subject Test: Math

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Question

Find the local maximum for the function .

Answer

To find the local max, you must find the first derivative, which is .

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 , giving you two critical points at and .

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.

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Question

Find the local maximum of the function.

Answer

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 is $nx^{n-1}$ . The derivative of the given function is

We must now set it equal to zero and factor.

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

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

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Question

Find the local maximum of the function.

Answer

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 is $nx^{n-1}$ . The derivative of the function is

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

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.

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.

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Question

Find the coordinate of the local maximum of the folowing function.

Answer

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 is $nx^{n-1}$ . Thus the derivative of the function is

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

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

So the local max is at x=1.

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Question

Find the minimum of the function:

Answer

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

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Question

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

Answer

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 $2+\frac{1}{3}x^5$ .

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Question

Let $f(x)=(1+\frac{x}{2})^{1/3}$

Find the the first three terms of the Taylor Series for centered at .

Answer

Using the formula of a binomial series centered at 0:

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

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

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

Then, we find the terms where,

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

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Question

Determine the convergence of the Taylor Series for at where $f(x)=\sum_{n=0}^{\infty }\frac{x^n}{n!}$ .

Answer

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

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

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Question

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

Answer

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.

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Question

Suppose that the derivative of a function, denoted , can be approximated by the third degree Taylor polynomial, centered at :

$f'(x) \approx 1 + (x-2) + \frac{4}{3} (x-2)^2 + \frac{8}{9}(x-2)^3$

If , find the third degree Taylor polynomial for centered at .

Answer

To get , we need to find the antiderivative of 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:

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

Since , substitute for the final .

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

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Question

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

Answer

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

.

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Question

Find the first two terms of the Taylor series about for the following function:

Answer

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

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

First, we must find the zeroth and first derivative of the function.

The zeroth derivative of a function is just the function itself, so we only have to find the first derivative:

The derivative was found using the following rule:

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

Now, write the first two terms of the sequence (n=0 and n=1):

$\frac{[2(3)+5](x-3)^0}{0!}+\frac{2(x-3)^1}{1!}=11+2(x-3)$

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Question

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

Answer

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.

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now use the above formula to write out the first three terms:

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

Simplified, this becomes

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