Differential Equations - AP Calculus AB

Card 0 of 2156

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

For the equation , graph the function, and identify where the local minima is.

Answer

By graphing the equation , we can see that there minimum at , and that the graph continues to rise in both directions around this point, so this must be a local minimum. We also know that the graph rises infinitely in both directions, so this must be the only local minimum.

Another way to identify the local minima is by taking the derivative of the function and setting it equal to zero.

Using the power rule,

$f(x)=x^n\rightarrow f'(x)=nx^{n-1}$ we find the derivative to be,

$y=x^2 \rightarrow y'=2\cdot x^{2-1}=2x$ .

From here we set the derivative equal to zero and solve for x. By doing this we will identify the critical values of the function

Now we will plug in the x value and find the corresponding y value in the original equation. We will also plug in an x value that is lower than the critical x value and a x value that is higher than the critical value to confirm whether we have a local minima or maxima.

Since both of the x values have a larger y value than the y value that corresponds to , we know that the minimum occurs at .

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Question

Find the local minimum of the function.

Answer

The points where the derivative of the function is equal to 0 are called critical points. They are either local maximums, local minimums, or do not exist. The derivative of is $nx^{n-1}$ . The derivative of the function is

.

We must now set it equal to zero and factor to solve.

Now we must plug in points to the left and right of the critical points into the derivative function to find the local min.

This means the function is increasing until it hits x=2, then it decreases until it hits x=4 and begins increasing again. This makes x=4 the local minumum

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Question

Find the coordinate of the local minumum of the following function.

Answer

The local maximums and minumums of a function are where the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative. Then we must set ot equal to 0 and solve. The derivative of is $nx^{n-1}$ .

The critical points are at the above two points. To find the minimum we must plug both back into the origianl function.

Thus the local min is at x=-2.

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Question

You are given the function . Find the minimum point of the function.

Answer

To find the minimum of a function, start by finding the critical points of that function, or points where the derivative is equal to zero. Use the power rule to find the derivative:

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Applying the power rule to the given equation, noting the constants in the first and second terms:

$\frac{d}{dx} f(x) = 4x +16$

Then check to see if the critical point is a maximum, minimum, or an inflection point by taking the second derivative, using the power rule once again.

$\frac{d^2}{dx^2}f(x) = 4$

Because the second derivative is positive, the critical point is a minimum.

To find the point where the minimum occurs, plug back into the original equation and solve for .

Therefore, the minimum is

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Question

A function is given by the equation

$f(x) = \frac{(x-4)^{2}}{8}+2x+3$ .

By graphing the derivative of , which value corresponds to the local minumum?

Answer

The local minimum of a function can be found by finding the derivative and graphing it. The point in which the x axis is crossed from below gives the x position where the local minimum is found. Taking the derivative:

$f(x) = \frac{(x-4)^{2}}{8}+2x+3$

$f'(x) = \frac{2(x-4)^{2-1}}{8}+2 = \frac{1}{4}(x-4)+2 = 0.25x+ 1$

The graph of the derivative is shown below:

As shown by the graph, the local minimum is found at x = -4.

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Question

The function f(x) is shown here in the graph

Without solving for the derivative, which of the following graphs is the graph of the derivative of , i.e the graph of ?

Answer

In order to determine the graph by inspection, there are key features to look for. The most important is the locations of the local maxima and minima in the graph of f(x). These points correspond to the x-intercepts in the graph of the derivative. Taking a look at the graph of f(x), you can see that the x intercepts on the graph of f'(x) will be located roughly at x = -3 and x = 4.5. Looking at the possible answers, the only two that could be graphs of f'(x) are these two:

and

The next step would then be to see which corresponds correctly to maxima and minima. Since the point at x = -3 is a local maximum, f(x) will increase up until the point at which it is maximum, then begin to drop. As seen in the positively oriented parabola, the rate of change of f(x) (the derivative) is positive up until it reaches x = -3. This means that f(x) was increasing, and indicates that this point was a local maximum. On the other hand, if you look at the graph on the left with the negatively oriented parabola, f'(x) is negative until it reaches the local maximum, which doesn't make sense, since that would mean it was decreasing up until the point and then increasing. This indicates a minima.

Since the point at x = -3 is a local maximum, the only graph that could be the derivative of f(x) is the positively oriented parabola.

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Question

Find the derivative of (5+3x)5.

Answer

We'll solve this using the chain rule.

Dx\[(5+3x)5\]

=5(5+3x)4 * Dx\[5+3x\]

=5(5+3x)4(3)

=15(5+3x)4

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Question

Find Dx\[sin(7x)\].

Answer

First, remember that Dx\[sin(x)\]=cos(x). Now we can solve the problem using the Chain Rule.

Dx\[sin(7x)\]

=cos(7x)Dx\[7x\]

=cos(7x)(7)

=7cos(7x)

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Question

Calculate fxxyz if f(x,y,z)=sin(4x+yz).

Answer

We can calculate this answer in steps. We start with differentiating in terms of the left most variable in "xxyz". So here we start by taking the derivative with respect to x.

First, fx= 4cos(4x+yz)

Then, fxx= -16sin(4x+yz)

fxxy= -16zcos(4x+yz)

Finally, fxxyz= -16cos(4x+yz) + 16yzsin(4x+yz)

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Question

Integrate $\small \int_{0}^{\pi } \cos x \hspace 2 dx$

Answer

$\frac{\mathrm{d} }{\mathrm{d} x}\sin x=cosx$

thus:

$\small \small \int_{0}^{\pi } \cos x \hspace 2 dx = \sin x \left]^\pi_0$
$\small \sin \pi - \sin 0 = 0 - 0 = 0$

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Question

Integrate :

$\small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx$

Answer

$\frac{\mathrm{d} }{\mathrm{d} x}\sec x=\sec x\tan x$

thus:
$\small \small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx = \sec x \left]^0_{-\frac{\pi}{4} }$
$\small = \sec 0 - \sec(-\frac{\pi}{4}) = 1 - \sqrt{2}$

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Question

Find the general solution, , to the differential equation

$\frac{dy}{dx}=30-2x$ .

Answer

We can use separation of variables to solve this problem since all of the "y-terms" are on one side and all of the "x-terms" are on the other side. The equation can be written as $\frac{dy}{dx}=30-2x\Rightarrow dy=(30-2x)dx$ .

Integrating both sides gives us $\int dy = \int (30-2x) dx\Rightarrow y(x) = 30x-x^2+C$ .

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Question

Consider ; by multiplying by both the left and the right hand sides can be swiftly integrated as

$\frac{1}{2}(y')^2 = F(y) + C$

where . So, for example, can be rewritten as:

$\frac{1}{2} [y'(x)]^2 = \frac{1}{3} [y(x)]^3 + C$ . We will use this trick on another simple case with an exact integral.

Use the technique above to find such that with and .

Hint: Once you use the above to simplify the expression to the form , you can solve it by moving into the denominator:

$\int dy/g(y) = \int dx$

Answer

As described in the problem, we are given

.

We can multiply both sides by :

Recognize the pattern of the chain rule in two different ways:

$\frac{1}{2}[(y')^2]' = [y^2]' + \frac{1}{2} [y^4]'$

This yields:

We use the initial conditions to solve for C, noticing that at and This means that C must be 1 above, which makes the right hand side a perfect square:

$\frac{dy}{dx} = \pm (1 + y^2)$

To see whether the + or - symbol is to be used, we see that the derivative starts out positive, so the positive square root is to be used. Then following the hint we can rewrite it as:

$\int \frac{dy}{1 + y^2} = \int dx$ ,

which we learned to solve by the trigonometric substitution, yielding:

$\tan^{-1} y = x + C$

Clearly and the fact that again gives us so

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Question

What are all the functions such that

?

Answer

Integrating once, we get:

$y'(x) = \int \frac{dx}{x} + C_1 = \ln x + C_1$

Integrating a second time gives:

$y(x) = \int \ln x ~ dx + C_1 ~ x + C$

We integrate the first term by parts using to get:

$y(x) = x ~ \ln x ~-~ \int x ~ \frac{dx}{x} ~+~ C_1 ~ x ~+~ C$

Canceling the x's we get:

$y(x) = x ~ \ln x ~+~ \left( C_1 - 1 \right ) x ~+~ C$

Defining gives the above form.

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Question

The Fibonacci numbers are defined as

$F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$

and are intimately tied to the golden ratios $\phi_\pm = (1 \pm \sqrt{5})/2$ , which solve the very similar equation

$\phi^n_\pm = \phi^{n - 1}_\pm + \phi^{n-2}_\pm$ .

The n'th derivatives of a function are defined as:

$y^{(0)}(x) = y(x), y^{(n)}(x) = \frac{d}{dx} y^{(n - 1)}(x)$

Find the Fibonacci function defined by:

$f^{(n)}(x) = f^{(n-1)}(x) + f^{(n-2)}(x)$

$f^{(0)}(0) = 0, f^{(1)}(0) = 1$

whose derivatives at 0 are therefore the Fibonacci numbers.

Answer

To solve $f^{(n)} = f^{(n-1)} + f^{(n-2)}$ , we ignore of the derivatives to get simply:

This can be solved by assuming an exponential function $y = C e^{k x}$ , which turns this expression into

,

which is solved by $k = \phi_\pm$ . Our general solution must take the form:

$f(x) = C_+ ~ e^{x \phi_+} + C_- ~ e^{x \phi_-}$

Plugging in our initial conditions and , we get:

$C_+ \ \phi_+ ~+~ C_- \ \phi_- = 1$

$C_+ = 1/(\phi_+ - \phi_-) = 1/(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}) = 1/\sqrt{5}$

Hence the answer is:

$f(x) = (e^{x \phi_+} - e^{x \phi_-}) / \sqrt{5}$

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Question

Find of the following equation:

Answer

First take the derivative and then solve when x=2.

To find the derivative use the power rule which states when,

the derivative is $f'(x)=nx^{n-1}$ .

Therefore the derivative of our function is:

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