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lesya692 [45]
2 years ago
15

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 5 + 54x − 2x3, [0, 4]

Mathematics
1 answer:
Mazyrski [523]2 years ago
5 0

Answer:

The absolute maximum and minimum of f(x) = 5 + 54\cdot x -2\cdot x^{3} on [0,4] are 113 and 5.

Step-by-step explanation:

Let be f(x) = 5 + 54\cdot x -2\cdot x^{3}, the first and second derivatives of the function are, respectively:

f'=54-6\cdot x^{2}

f''=-12\cdot x

Now, let equalize the first derivative to zero and solve the resulting expression:

54-6\cdot x^{2} = 0

x^{2} = 9

x =\pm 3

According to the given interval, only x=3 is a valid outcome. Lastly, this is evaluated in the second derivative expression:

f''=-12\cdot(3)

f'' = -36

x = 3 leads to an absolute maximum.

f(3) = 5 + 54\cdot (3) -2\cdot (3)^{3}

f(3) = 113

The absolute minimum is determined by evaluating at each extreme of the interval:

x = 0

f(0) = 5 + 54\cdot (0) -2\cdot (0)^{3}

f(0) = 5

x = 4

f(4) = 5 + 54\cdot (4) -2\cdot (4)^{3}

f(4) = 93

The absolute maximum and minimum of f(x) = 5 + 54\cdot x -2\cdot x^{3} on [0,4] are 113 and 5.

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2^t =38

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t log (2) = log (38)

divide by log 2

t = log(38)/log (2)

t≈5.2479


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3 years ago
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Bingel [31]
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4 0
2 years ago
For the equation y= x + 5, find the value of y when
Nina [5.8K]

Answer:

a) 3 b) 5 c) 7 d) 9

Step-by-step explanation:

For this, you want to replace x in the equation y=x+5 with each of the values listed.

For the first one, -2, the equation becomes y=-2+5, which is solved to y=3.

For the second one, 0, the equation becomes y=0+5, which is solved to y=5.

For the third one, 2, the equation becomes y=2+5, which is solved to y=7.

For the last one, 4, the equation becomes y=4+5, which is solved to y=9.

**This content involves solving algebraic equations with a known variable, which you may wish to revise. I'm always happy to help!

5 0
2 years ago
Which expression is equivalent to (x^4/3 x^2/3)^1/3?
KATRIN_1 [288]

Answer: x^{\frac{2}{3} }

Step-by-step explanation:

We have several properties of exponents in use here. The two that are used are:

(x^{a})(x^{b}) = x^{a + b} <em>(Exponents with the same base that are being multiplied together can have the exponents added)</em>

(x^{a})^{b} = x^{(a)(b)} <em>(A base raised to a power, and then raised to another power means that you can multiply the exponents to get the same result as doing inside operations and then outside operations)</em>

<em />

Let's apply it!

First, let's simplify what's inside the parenthesis.

x^{\frac{4}{3} } x^{\frac{2}{3} } <em>(Remember, they have the same base of "x", so we can add the exponents)</em>

x^{\frac{4}{3} + \frac{2}{3} } = x^{\frac{6}{3} } = x^{2}

Now we have (x^{2})^{\frac{1}{3} }. Let's use the second rule.

(x^{2})^{\frac{1}{3} } = x^{\frac{2}{3} }

Hope this helps! :^)

7 0
3 years ago
Solve for x.<br>-3+85-5=-8<br>A. -2.<br>B.-1<br>C.-3/4<br>D.0​
posledela

Answer:

-2

Step-by-step explanation:

-3+8x-5=-8

8x = -16

x = -2

5 0
3 years ago
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