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Blababa [14]
2 years ago
5

Brainliest if correct Solve: 8( 4 - 2^{6} divided by 4)

Mathematics
1 answer:
polet [3.4K]2 years ago
6 0

\huge\text{Hey there!}

\mathsf{\dfrac{8(4 - 2^6)}{4}}

\mathsf{2^6}\\\mathsf{\rightarrow2\times2\times2\times2\times2\times2}\\\mathsf{\rightarrow 4\times4\times4}\\\mathsf{\rightarrow 16\times4}\\\mathsf{\rightarrow \bf 64}

\mathsf{= \dfrac{8(4 - \bf 64)}{4}}

\mathsf{4 - 64}\\\mathsf{\rightarrow \bf -60}

\mathsf{= \dfrac{8(\bf -60)}{4}}

\mathsf{8(-60)}\\\mathsf{\rightarrow \bf -480}

\mathsf{= \dfrac{-480}{4}}\\\\\mathsf{\rightarrow \dfrac{-480\div4}{4\div4}}\\\\\mathsf{\rightarrow- \dfrac{\bf 120}{1}}\\\\\mathsf{\rightarrow \bold{-120}\div 1}}\\\\\mathsf{\rightarrow \bf -120}

\huge\text{Therefore, your answer is: \boxed{\mathsf{-120}}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy your day!}

~\frak{Amphitrite1040:)}

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Claire says the area of the square face is 8 m2. Is she correct? No, the face’s dimensions are 4 m by 7 m, so the area is 14 m2.
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Mr. Moses baked 11 dozen cookies for his class.
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3 years ago
What is the upper bound of the function f(x)=4x4−2x3+x−5?
inessss [21]

Answer:

(no global maxima found)

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = 4 x^4 - 2 x^3 + x - 5

Hint: | Global extrema of f(x) can occur only at the critical points or the endpoints of the domain.

Find the critical points of f(x):

Compute the critical points of 4 x^4 - 2 x^3 + x - 5

Hint: | To find critical points, find where f'(x) is zero or where f'(x) does not exist. First, find the derivative of 4 x^4 - 2 x^3 + x - 5.

To find all critical points, first compute f'(x):

d/( dx)(4 x^4 - 2 x^3 + x - 5) = 16 x^3 - 6 x^2 + 1:

f'(x) = 16 x^3 - 6 x^2 + 1

Hint: | Find where f'(x) is zero by solving 16 x^3 - 6 x^2 + 1 = 0.

Solving 16 x^3 - 6 x^2 + 1 = 0 yields x≈-0.303504:

x = -0.303504

Hint: | Find where f'(x) = 16 x^3 - 6 x^2 + 1 does not exist.

f'(x) exists everywhere:

16 x^3 - 6 x^2 + 1 exists everywhere

Hint: | Collect results.

The only critical point of 4 x^4 - 2 x^3 + x - 5 is at x = -0.303504:

x = -0.303504

Hint: | Determine the endpoints of the domain of f(x).

The domain of 4 x^4 - 2 x^3 + x - 5 is R:

The endpoints of R are x = -∞ and ∞

Hint: | Evaluate f(x) at the critical points and at the endpoints of the domain, taking limits if necessary.

Evaluate 4 x^4 - 2 x^3 + x - 5 at x = -∞, -0.303504 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-0.303504 | -5.21365

∞ | ∞

Hint: | Determine the largest and smallest values that f achieves at these points.

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-0.303504 | -5.21365 | global min

∞ | ∞ | global max

Hint: | Finally, remove the endpoints of the domain where f(x) is not defined.

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-0.303504 | -5.21365 | global min

Hint: | Summarize the results.

f(x) = 4 x^4 - 2 x^3 + x - 5 has one global minimum:

Answer: f(x) has a global minimum at x = -0.303504

5 0
3 years ago
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