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Alona [7]
3 years ago
12

The graph of y = f ′(x), the derivative of f(x), is shown below. Given f(2) = 8, evaluate f(–2).

Mathematics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:  Third option is correct.

Step-by-step explanation:

Since we have given that

The graph of the derivative of f(x) , say, y = f'(x), is symmetry about the origin.

so, it is an odd function.

So, f(x) must be an even function.

f(x)=f(-x)\\\\f(2)=f(-2)=8

So, the value of f(-2) is equal to 8.

Hence, Third option is correct.

You might be interested in
Consider the integral Integral from 0 to 1 e Superscript 6 x Baseline dx with nequals 25 . a. Find the trapezoid rule approximat
photoshop1234 [79]

Answer:

a.

With n = 25, \int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549

With n = 50, \int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594

b. \int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943

c.

The absolute error in the trapezoid rule is 0.08047

The absolute error in the Simpson's rule is 0.00008

Step-by-step explanation:

a. To approximate the integral \int_{0}^{1}e^{6 x}\ dx using n = 25 with the trapezoid rule you must:

The trapezoidal rule states that

\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)

where \Delta{x}=\frac{b-a}{n}

We have that a = 0, b = 1, n = 25.

Therefore,

\Delta{x}=\frac{1-0}{25}=\frac{1}{25}

We need to divide the interval [0,1] into n = 25 sub-intervals of length \Delta{x}=\frac{1}{25}, with the following endpoints:

a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b

Now, we just evaluate the function at these endpoints:

f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1

2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281

2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579

...

2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701

f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735

Applying the trapezoid rule formula we get

\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549

  • To approximate the integral \int_{0}^{1}e^{6 x}\ dx using n = 50 with the trapezoid rule you must:

We have that a = 0, b = 1, n = 50.

Therefore,

\Delta{x}=\frac{1-0}{50}=\frac{1}{50}

We need to divide the interval [0,1] into n = 50 sub-intervals of length \Delta{x}=\frac{1}{50}, with the following endpoints:

a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b

Now, we just evaluate the function at these endpoints:

f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1

2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875

2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281

...

2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705

f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735

Applying the trapezoid rule formula we get

\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594

b. To approximate the integral \int_{0}^{1}e^{6 x}\ dx using 2n with the Simpson's rule you must:

The Simpson's rule states that

\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)

where \Delta{x}=\frac{b-a}{n}

We have that a = 0, b = 1, n = 50

Therefore,

\Delta{x}=\frac{1-0}{50}=\frac{1}{50}

We need to divide the interval [0,1] into n = 50 sub-intervals of length \Delta{x}=\frac{1}{50}, with the following endpoints:

a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b

Now, we just evaluate the function at these endpoints:

f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1

4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175

2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281

...

4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541

f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735

Applying the Simpson's rule formula we get

\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943

c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by |A-B|

The absolute error in the trapezoid rule is

The calculated value is

\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225

and our estimate is 67.1519320308594

Thus, the absolute error is given by

|67.0714655821225-67.1519320308594|=0.08047

The absolute error in the Simpson's rule is

|67.0714655821225-67.0715427161943|=0.00008

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Help!!<br><br> Simplify the fraction 56/64 as much as possible.
STatiana [176]

Answer:

simplest form is 7/8

Step-by-step explanation:

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Answer:

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6 0
2 years ago
What causes an electric charge to jump from your hand to a metal object?
meriva

Answer:

Charges are both negative and positive as such negative charges are drawn to positive charges and the same with positive being drawn to negative

this can be explained because the human body has natural medals in that attract negative and positive charges just like metal.

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Part B what are the probabilities of each outcome in the sample space? select all that apply
zepelin [54]

Answer:

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Step-by-step explanation:

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<em />

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