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Aleksandr-060686 [28]
3 years ago
14

The figure below shows the graph of f ', the derivative of the function f, on the closed interval from x = -2 to x = 6. The grap

h of the derivative has horizontal tangent lines at x = 2 and x = 4.
Find the x-value where f attains its absolute maximum value on the closed interval from x = -2 to x = 6.

Mathematics
1 answer:
wolverine [178]3 years ago
6 0

Answer:

x = -2

Step-by-step explanation:

From x = -2 to x = 5, f' is negative.  That means f is decreasing.

From x = 5 to x = 6, f' is positive.  That means f is increasing.

The negative area (between x = -2 and x = 5) is larger than the positive area area (between x = 5 and x = 6).  That means f decreases more than it increases.

So f is an absolute maximum at x = -2.

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Answer:

9 times. 72/8=t or 72=8t.

Explanation:

I'm afraid the answer choices are slightly mixed up, but acceptable answers would probably include 72/8=t or 72=8t. Simply put the two numbers into a fraction (72/8) to get how many times larger or smaller the thing is. In this case, it would be 9 times as many worms.

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3 years ago
Read 2 more answers
Find maclaurin series
Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

The first 3 terms of the series are

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and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}

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Answer:

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sqrt(46)

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So the sqrt(46) is between 6 and 7

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