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3 years ago

What are the zeros of this function? look at grapha.b.c.d.

Mathematics

1 answer:

fredd [130]3 years ago

4 0

The zeros of the function occur when the graph meet the x-axis so in this case the zeros occur at B ( 0 and 5)

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X= <br> A. 115.5<br> B. 137<br> C. 180

zhuklara [117]

<u>Answer:</u>

The correct answer option is C. 180

<u>Step-by-step explanation:</u>

We are given a figure of a circle with two known angles measuring 137 degrees and 94 degrees and we are to find the value of x.

Finding x:

$137 = \frac{1}{2} (\angle x + 94)$

$137 = \frac{1}{2} \angle x + 47$

$\frac{1}{2} \angle x = 137 - 47$

$\frac{1}{2} \angle x = 90
$

$\angle x = 90 \times 2&# 1 0$

x = 180

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3 years ago

Evaluate x ➗ y, if x = 29.665 and y = 6.98

Oduvanchick [21]

Answer:

4.25

Step-by-step explanation:

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3 years ago

If you are picking a number between 1-20 what is the probability that you will pick a multiple of two or a number greater than 1

Alisiya [41]

Multiple of two has a 50 percent chance. greater than 15 is 25 percent. if both it's 10 percent

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4 years ago

Read 2 more answers

What should be done first to simplify the expression?

Olin [163]

Answer:

add 3+3

Step-by-step explanation:

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4 years ago

In Exercises 11-18, use analytic methods to find the extreme values of the function on the interval and where they occur. Identi

Colt1911 [192]

Answer:

Absolute maximum of 1 at x = pi/4 ; $(\frac{\pi}{4}, \ 1)$

Absolute minimum of -1 at x = 5pi/4 ; $(\frac{5\pi}{4} , \ -1)$

Local maximum of √2/2 at x = 0 ; $(0, \ \frac{\sqrt{2} }{2} )$

Local minimum of 0 at x = 7pi/4 ; $(\frac{7\pi}{4}, \ 0)$

No critical points that are not stationary points.

Step-by-step explanation:

$f(x)=sin(x+\frac{\pi}{4} ), \ 0 \leq x\leq \frac{7 \pi}{4}$

<h2>Take Derivative of f(x):</h2>

Let's start by taking the derivative of the function.

Use the power rule and the chain rule to take the derivative of f(x).

$f'(x)=\frac{d}{dx} [sin(x+\frac{\pi}{4})] \times \frac{d}{dx} (x+\frac{\pi}{4})$

The derivative of sin(x) is cos(x), so we can write this as:

$f'(x)=cos(x+\frac{\pi}{4})\times \frac{d}{dx} (x+\frac{\pi}{4})$

Now, we can apply the power rule to x + pi/4.

$f'(x)=cos(x+\frac{\pi}{4} ) \times 1$

$f'(x)=cos(x+\frac{\pi}{4} )$

<h2>Critical Points: Set f'(x) = 0</h2>

Now that we have the first derivative of $f(x)=sin(x+\frac{\pi}{4})$ , let's set the first derivative to 0 to find the critical points of this function.

$0=cos(x+\frac{\pi}{4})$

Take the inverse cosine of both sides of the equation.

$cos^-^1(0) = cos^-^1[cos(x+\frac{\pi}{4})]$

Inverse cosine and cosine cancel out, leaving us with x + pi/4. The inverse cosine of 0 is equal to 90 degrees, which is the same as pi/2.

$\frac{\pi}{2} = x +\frac{\pi}{4}$

Solve for x to find the critical points of f(x). Subtract pi/4 from both sides of the equation, and move x to the left using the symmetric property of equality.

$x=\frac{\pi}{2}- \frac{\pi}{4}$

$x=\frac{2 \pi}{4}-\frac{\pi}{4}$
$x=\frac{\pi}{4}$

Since we are given the domain of the function, let's use the period of sin to find our other critical point: 5pi/4. This is equivalent to pi/4. Therefore, our critical points are:

$\frac{\pi}{4}, \frac{5 \pi}{4}$

<h2>Sign Chart(?):</h2>

Since this is a sine graph, we don't need to create a sign chart to check if the critical values are, in fact, extreme values since there are many absolute maximums and absolute minimums on the sine graph.

There will always be either an absolute maximum or an absolute minimum at the critical values where the first derivative is equal to 0, because this is where the sine graph curves and forms these.

Therefore, we can plug the critical values into the original function f(x) in order to find the value at which these extreme values occur. We also need to plug in the endpoints of the function, which are the domain restrictions.

Let's plug in the critical point values and endpoint values into the function f(x) to find where the extreme values occur on the graph of this function.

<h2>Critical Point Values:</h2>

$f(\frac{\pi}{4} )=sin(\frac{\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{\pi}{4} )=sin(\frac{2\pi}{4}) \\ f(\frac{\pi}{4} )=sin(\frac{\pi}{2}) \\ f(\frac{\pi}{4} )=1$

There is a maximum value of 1 at x = pi/4.

$f(\frac{5\pi}{4} )=sin(\frac{5\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{5\pi}{4} )=sin(\frac{6\pi}{4}) \\ f(\frac{5\pi}{4}) = sin(\frac{3\pi}{2}) \\ f(\frac{5\pi}{4} )=-1$

There is a minimum value of -1 at x = 5pi/4.

<h2>Endpoint Values:</h2>

$f(0) = sin((0) + \frac{\pi}{4}) \\ f(0) = sin(\frac{\pi}{4}) \\ f(0) = \frac{\sqrt{2} }{2}$

There is a maximum value of √2/2 at x = 0.

$f(\frac{7\pi}{4} ) =sin(\frac{7\pi}{4} +\frac{\pi}{4}) \\ f(\frac{7\pi}{4} ) =sin(\frac{8\pi}{4}) \\ f(\frac{7\pi}{4} ) =sin(2\pi) \\ f(\frac{7\pi}{4} ) =0$

There is a minimum value of 0 at x = 7pi/4.

We need to first compare the critical point values and then compare the endpoint values to determine whether they are maximum or minimums.

<h2>Stationary Points:</h2>

A critical point is called a stationary point if f'(x) = 0.

Since f'(x) is zero at both of the critical points, there are no critical points that are not stationary points.

6 0

3 years ago