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2 years ago

F(x) = (x - 1) (4x + 3) (3x - 8) has zeros at x = 3/4, x= 1, and x = 8/3.

Mathematics

1 answer:

inn [45]2 years ago

8 0

Using the given roots, it is found that on the interval $-\frac{3}{4} < x < 1$ , the function is positive.

The function is:

$f(x) = (x - 1)(4x + 3)(3x - 8)$

The zeroes are:

$x = -\frac{3}{4}, x = 1, x = \frac{8}{3}$

Between two zeroes, the function has the same sign, hence, to find the sign on the interval $-\frac{3}{4} < x < 1$ , we find the numeric value of the function of a value of x on the interval, for example f(0), for x = 0.

Then:

$f(0) = (0 - 1)[4(0) + 3][3(0) - 8] = (-1)(3)(-8) = 24$

Since f(0) is positive, the function is positive on the interval $-\frac{3}{4} < x < 1$ .

A similar problem, in which it is desired to find the sign of the function, is given at brainly.com/question/4787253

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2 years ago

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tankabanditka [31]

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2 years ago

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2 years ago

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Answer:

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Step-by-step explanation:

I made this into 3 rectangles.

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Figure 2:

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The 7 is from the 10 m side, but I've taken 4m off from the overlapping square again.

Remaining Area:

If you extend the lines into figures 1 and 2 from the top left corner and bottom right corner vertically, you will get a rectangle that is (11 m) x (4 m). This is not yet accounted for.

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2 years ago

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soldier1979 [14.2K]

Answer:

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Step-by-step explanation:

Given that :

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2 years ago