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2 years ago

Given the function f(x)= x^4 - 4x^3 -2, determine the absolute minimum value of ff on the closed interval [-2, 4].

Mathematics

1 answer:

kobusy [5.1K]2 years ago

4 0

Answer:

Absolute Minimum: (3,−29)

Step-by-step explanation:

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Please help on this ill give braneliest

horsena [70]

Answer:

-2.5

Step-by-step explanation:

25/-10 = -2.5

20/-8 = 2.5

10/-4 = 2.5

-5/2 = 2.5

so the answer is -2.5

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2 years ago

Pythagorean theorem to find the length of the hypotenuse in the triangle 16 30

pishuonlain [190]

Answer:

the answer is 34.

Step-by-step explanation:

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2 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

3 0

2 years ago

❤Please help sweeties!❤

Alona [7]

Answer:

satisfy both equations

Step-by-step explanation:

4 0

2 years ago

40 points plz help if u can Which function has no Zeroes? Select all that apply.

Damm [24]

<h3>2 Answers: Choice C and choice D</h3>

y = csc(x) and y = sec(x)

==========================================================

Explanation:

The term "zeroes" in this case is the same as "roots" and "x intercepts". Any root is of the form (k, 0), where k is some real number. A root always occurs when y = 0.

Use GeoGebra, Desmos, or any graphing tool you prefer. If you graphed y = cos(x), you'll see that the curve crosses the x axis infinitely many times. Therefore, it has infinitely many roots. We can cross choice A off the list.

The same applies to...

y = cot(x)
y = sin(x)
y = tan(x)

So we can rule out choices B, E and F.

Only choice C and D have graphs that do not have any x intercepts at all.

------------

If you're curious why csc doesn't have any roots, consider the fact that

csc(x) = 1/sin(x)

and ask yourself "when is that fraction equal to zero?". The answer is "never" because the numerator is always 1, and the denominator cannot be zero. If the denominator were zero, then we'd have a division by zero error. So that's why csc(x) can't ever be zero. The same applies to sec(x) as well.

sec(x) = 1/cos(x)

7 0

2 years ago