Graphing Functions - AP Calculus AB

Card 0 of 3836

Question

On which interval is the function shown in the above graph strictly decreasing?

Answer

A function is strictly decreasing on an inverval if, for any in the interval, (i.e the slope is always less than zero)

Interval E is the only interval on which the function shows this property.

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Question

Let $f(x)=\frac{4}{3}x^3-9x$ .

On which open interval(s) is decreasing?

Answer

is decreasing on intervals where .

First, differentiate .

Then, find the values for x for which the derivative is negative by solving

$x=\frac{3}{2}\textrm{ or} \frac{-3}{2}$ .

Next, test the intervals.

$\left(-\infty, -\frac{3}{2}\right)$

$\left(-\frac{3}{2}, \frac{3}{2}\right)$

$\left(\frac{3}{2}, \infty\right)$

Test them by substituting values for x:

$\left(-\infty, -\frac{3}{2}\right)$

Substitute -2,

.

The function is increasing on this interval since the derivative is positive on this interval.

$\left(-\frac{3}{2}, \frac{3}{2}\right)$

Substitute 0,

.

The function is deecreasing on this interval since the derivative is negative on this interval.

$\left(\frac{3}{2}, \infty\right)$

Substitute 2,

.

The function is increasing on this interval since the derivative is positive on this interval.

Thus, $\left(-\frac{3}{2}, \frac{3}{2}\right)$ is the only interval on which the function is decreasing.

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Question

For which values of is the function $f(x)=x^4 -\frac{4}{3}x^3 -12x^2 +3$ decreasing?

Answer

To determine where the function is decreasing, differentiate it:

What we are interested in are the points where . To determine these points, factor the equation:

this has solutions at

This splits the graph into 4 regions, and we can test points in each to determine if is greater than or less than 0. If it is less than zero, the function is decreasing.

$- \infty < x < -2: f'(-3)=-72$ negative/decreasing

positive/increasing

negative/decreasing

$3 < x < \infty: f'(4)=96$ positive/increasing

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Question

For which values of is the function $f(x)= \frac{4}{3}x^3 + 9x^2 -10x$ decreasing?

Answer

The function is increasing where . To determine the regions where this is true, first take the derivative of :

.

To figure out where this is less than zero, factor and set it equal to zero. This will split the function into intervals where we can test points.

This has solutions at $x=-5, \frac{1}{2}$ .

Test a point in each region to see if the function is increasing or decreasing:

$- \infty < x < -5: f'(-6)=26$ positive/increasing

$-5 < x < \frac{1}{2}: f'(0)=-10$ negative/decreasing

$\frac{1}{2} < x < \infty: f'(1)=12$ positive/increasing

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Question

The picture below shows the graph of some function,

On which interval of is the function decreasing?

Answer

A function is decreasing when it has a negative slope. Graphically, this is a region of the curve where the curve decreases as increases.

On interval D, the curve shows this trait. The curve does not show this trait on any other interval.

Therefore, interval D is the answer.

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Question

Find the intervals on which the function is decreasing:

Answer

To determine the intervals on which the function is decreasing, we must find the intervals on which the first derivative of the function is negative.

The first derivative of the function is

and we used the following rule:

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

Now, we find the point(s) at which the first derivative equals zero - the critical value(s):

$c=\frac{\sqrt{3}}{3}$

Now, we make our intervals on which we see whether the first derivative is positive or negative:

$\left(-\infty , \frac{\sqrt{3}}{3}\right), \left(\frac{\sqrt{3}}{3}, \infty \right)$

On the first interval, the first derivative is negative, while on the second interval it is positive. Thus, the first interval is the one where the function is decreasing.

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Question

Given $f'(x) = 2e^{2x} - 4$ , find the interval over which is decreasing.

Answer

To find when a function is decreasing, we must first find where the critical points of the function are. Since we are given the derivative , we start by first setting the derivative equal to and solving for .

$2e^{2x} - 4 = 0$

$2e^{2x}=4$

$e^{2x} = 2$

$x = \frac{ln(2)}{2}$

This is our critical point. To evaluate where the function is decreasing, we must check the sign of the derivative on both sides of the critical point. Because $\frac{ln(2)}{2} > 0$ , we can check the left side of the critical point by plugging into .

Because , the function is decreasing on the left side of the critical point, on the interval $(-\infty, \frac{ln(2)}{2})$ .

Now we must check the right side of the critical point. Because $\frac{ln(2)}{2} < 1$ , we can check the right side of the critical point by plugging into .

Because , the function is increasing on the right side of the critical point, meaning the only interval on which the function is decreasing is $(-\infty, \frac{ln(2)}{2})$ .

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Question

Determine the intervals on which the given function is decreasing:

Answer

To determine the intervals on which the function is decreasing, we must find the intervals on which the function's first derivative is negative. To do this, we must find the first derivative and the critical value(s) at which the first derivative is equal to zero:

The derivative was found using the following rule:

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

Now, setting the first derivative equal to zero, we get

So, now we can make our intervals to be analyzed (is the first derivative positive or negative on the interval?), in which c is the upper and lower bound:

$(-\infty, 0), (0, \infty)$

Note that at c the first derivative is neither positive nor negative.

On the first interval, the first derivative is always negative, so the function is always decreasing on this interval. On the second interval, the first derivative is always positive, therefore the function is increasing on this interval.

We are concerned with the interval where the function is decreasing, so $(-\infty , 0)$ is our answer.

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Question

Find the interval(s) where the function is decreasing.

Answer

To find the intervals where the function decreases, apply the first derivative test. Find the derivative, set equal to , and solve to find local extrema.

$\small f '(x) = 3x^2 + 8x + 4 = 0$

$\small (3x + 2)(x + 2) = 0$

So $\small x = -\frac{2}{3}$ or $\small x = -2$ .

Next, test points in each of the intervals delineated by the potential local extrema. For example:

$\small f '(-3) = 7$ , so the function increases to the left of $\small x = -2$ .

$\small f '(-1) = -1$ , so the function decreases on the interval $\small \left [-2, -\frac{2}{3} \right]$

$\small f '(0) = 4$ , so the function increases to the right of $\small x = -\frac{2}{3}$ .

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Question

Find the intervals on which the given function is decreasing:

Answer

To determine where the function is decreasing, we must find the intervals on which the first derivative of the function is negative.

First, we must find the first derivative:

The first derivative was found using the following rule:

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

Now, we set the first derivative equal to zero and solve for the critical values - values at which the first derivative is equal to zero:

Now, using the critical values, we create the intervals in which we see whether the first derivative is positive or negative:

$(-\infty, -2), (-2, 2), (2, \infty)$

On the first interval, the derivative is positive, on the second it is negative, and on the third interval it is positive (we simply plug in any point in the interval into the first derivative function and check the sign).

So, the function is decreasing on the second interval .

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Question

Find the interval(s) on which the function is decreasing:

Answer

To determine the intervals on which the function is decreasing, we must find the intervals on which the first derivative of the function is negative.

First, we find the first derivative:

It was found using the following rule:
$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we must find the critical values - values at which the first derivative is equal to zero:

Now, using the critical values as upper and lower bounds, we create the intervals on which we determine the sign of the first derivative:

$(-\infty, 0), (0, \infty)$

On the first interval, the first derivative is always positive, and on the second interval, the first derivative is always positive. (Simply plug in any point in the interval into the first derivative function and check the sign.) Thus, the function is never decreasing.

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Question

Find the interval in which the following function is decreasing.

Answer

To find decreasing intervals, you must find when the first derivative is less than 0. Differentiate using the power rule:

$(x^n)'=nx^{n-1}$

Thus,

Since 2 is never negative, our first derivative is never negative. Therefore, our function is never decreasing.

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Question

One can find local (or absolute) maximum and minimum values of a differentiable function using calculus. Upon taking a function's derrivative, there may be points at which the derrivative is zero. These points indicate critical points (local or absolute max, min or, sometimes, an inflection point).

Where does the function $f(x)= \ln (\sin x)+x$ achieve a local maximum when $\pi < x < \frac{3\pi}{2}$ ?

Answer

Take the derivative of the function:

$f'(x)=\frac{d}{dx} [\ln (\sin x)+x]$

$\frac{dy}{dx}=\frac{1}{\sin x}*\cos x +1$

$\frac{dy}{dx}=\cot x+1$

To find critical points, we set dy/dx = 0

$0=\cot x+1$

or, equivalently,

$\cot x=-1$

The first couple points at which this is true are

$x= \frac{3 \pi}{4} , x= \frac{5 \pi}{4}$ .

These are both local maxima, but only $x=\frac{5 \pi}{4}$ fulfils the condition

$\pi < x < \frac{3\pi}{2}$ . Hence, $x=\frac{5 \pi}{4}$ is our answer.

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Question

Find the area bounded by the curve in the first quadrant.

Answer

The curve is in quadrant one over the interval , which are the bounds of integration. To see this, note that the x-intercepts are 0 and 2 and the parabola opens downward.

The definite integral below is solved by taking the antiderivative of each term of the given polynomial function, evaluating this antiderivative at the bounds of integration, and subtracting the values.

For this particular integral use the rule, $\int x^n=\frac{x^{n+1}}{n+1}$ to solve.

$\int_{0}^{2}(2x-x^2)dx=(x^2-\frac{1}{3}x^3)|^2_0=(2^2-\frac{1}{3}\cdot2^3)-(0^2-\frac{1}{3}\cdot0^3)=\frac{4}{3}$

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Question

Tessie kicks a bean bag into the air. Its height at a given time can be given by the following equation:

$h = 1+32t - 5t^{2}$

At what time will the bean bag reach its maximum height?

Answer

To find the time at which the bean bag reaches its maximum height, we need to find the time when the velocity of the bean bag is zero.

We can find this by taking the derivative of the height position function with respect to time:

Setting this equal to 0, we can then solve for t:

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Question

The postion of a feather in a windstorm is given by the following equation:

$x(t)= \frac{t^{3}}{3}+\frac{t^{2}}{2}-6t$

Determine when an extrema in the feather's position occurs and state whether it is a local minima or maxima.

Answer

The first step into finding when the extrema of a function occurs is to take the derivative and set it equal to zero:

$x(t)= \frac{t^{3}}{3}+\frac{t^{2}}{2}-6t$

$x'(t) =0= t^{2}+t-6$

To solve the right side of the equation, we'll need to find its roots. we may either use the quadratic formula:

$\frac{-1\pm \sqrt{1-4(1)(-6))}}{2}$

or see that the equation factors into:

Either way, the only nonnegative root is 2.

To see whether this a local minima or maxima, we'll take the second derivative of our equation and plug in this value of 2:

Since the second derivative is positive, we know that this represents a local minima.

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Question

Function gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Which of the following ordered pairs are the coordinates of a critical point of ?

Answer

Recall that a function has critcal points where its first derivative is equal to zero or undefined.

So, given , we need to find v'(t)

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

$\small \small \small v'(t)=5t^4+12t^3-\frac{14t}{3}$

Where is this function equal to zero? t=0 for one. We can find the others, but we really just need one. Plug in t=0 into our original equation to find the point (0,0) Which is in this case a saddle point.

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Question

At which point does a local maxima appear in the following function?

Answer

A local max will occur when the function changes from increasing to decreasing. This means that the derivative of the function will change from positive to negative.

First step is to find the derivative.

Find the critical points (when is or undefined).

$0=3x^2+4x, x=0,-\frac{4}{3}$

Next, find at which of these two values changes from positive to negative. Plug in a value in each of the regions into .

The regions to be tested are $\left(-\infty,-\frac{4}3\right)$ , $\left(-\frac{4}3,0\right)$ , and $(0,\infty)$ .

A value in the first region, such as , gives a positive number, and a value in the second range gives a negative number, meaning that $x=-\frac{4}{3}$ must be the point where the max occurs.

To find what the coordinate of this point, plug in $x=-\frac{4}{3}$ in to , not , to get .

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Question

What is the maximum value of the function on the interval ?

Answer

First, we need to find the critical points of the function by taking .

This is the derivative of a polynomial, so you can operate term by term.

This gives us,

.

Solving for by factoring, we get

.

This gives us critical values of 0 and $-\frac{2}{9}$ . Since we are operating on the interval , we make sure our endpoints are included and exclude critical values outside this interval. Now we know the maximum could either occur at or . As the function is decreasing, we know at the max occurs at and that that value is .

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Question

Consider the family of curves given by $f(x)=\frac{ax}{1+bx^2}$ with . If is a local maximum, determine and .

Answer

Since a local maximum occurs at , this tells us two pieces of information: the derivative of must be zero at and the point must lie on the graph of this function. Hence we must solve the following two equations:

.

From the first equation we get $1=\frac{a}{1+b}$ or .

To find the derivative we apply the quotient rule

$f'(x)=\frac{(1+bx^2)a-ax(2bx)}{(1+bx^2)^2}=\frac{a(1-bx^2)}{(1+bx^2)^2}$ .

Solving , we get . Plugging this into the expression for gives .

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