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Y_Kistochka [10]
2 years ago
12

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 <u>numeric value of the function of a value of x on the interval</u>, 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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