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Karo-lina-s [1.5K]
3 years ago
8

Help. I got it wrong and have no clue how to get to the right answer!

Mathematics
1 answer:
natima [27]3 years ago
7 0
\bf 2x+\sqrt{4x^2}+\sqrt[3]{8x^3}\qquad 
\begin{cases}
4\to 2^2\\
8\to 2^3
\end{cases}\qquad thus
\\\\\\
2x+\sqrt{2^2x^2}+\sqrt[3]{2^3x^3}\implies 2x+\sqrt{(2x)^2}+\sqrt[3]{(2x)^3}
\\\\\\
2x+2x+2x=\boxed{6x}
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Answer:  C) Sometimes positive; sometimes negative

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

Explanation:

Pick a value between x = -1 and x = 0. Let's say we go for x = -0.5

Plug this into f(x)

f(x) = x(x+3)(x+1)(x-4)

f(-0.5) = -0.5(-0.5+3)(-0.5+1)(-0.5-4)

f(-0.5) = -0.5(2.5)(0.5)(-4.5)

f(-0.5) = 2.8125

We get a positive value.

This shows that f(x) is positive on the region of -1 < x < 0

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

Now pick a value between x = 0 and x = 4. I'll use x = 1

f(x) = x(x+3)(x+1)(x-4)

f(1) = 1(1+3)(1+1)(1-4)

f(1) = 1(4)(2)(-3)

f(1) = -24

Therefore, f(x) is negative on the interval 0 < x < 4

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

In short, f(x) is both positive and negative on the interval -1 < x < 4

It's positive when -1 < x < 0

And it's negative when 0 < x < 4

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