Where does the function f have a zero? (more than one can apply)

Mathematics

1 answer:

alekssr [168]4 years ago

3 0

Answers:
Choice A(-2,-1)
Choice B(-1,0)
Choice E(2,3)

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Explanation:

The idea is to look for sign changes in f(x). Recall that y = f(x). If the sign of y changes, then y = 0 must be somewhere in the interval if we are to have a continuous function.

The sign changes happen when we go from...
y = -9 to y = +8 (row 2 to row 3)
y = +8 to y = -5 (row 3 to row 4)
y = -1 to y = +196 (bottom two rows)

Those three sign changes correspond to the following interval notations for x
(-2,-1)
(0,1)
(2,3)
in that exact order

Meaning for instance, that a root must be somewhere in the interval -2 < x < -1 since we have a sign change when we transition from f(-2) to f(-1)

So that's why the answers are choices A, B, & E

Something like the interval notation (1,2) will not work because y = -18 does not change sign to y = -1, which is why D is not an answer. Choice C is similar.

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A food packet is dropped from a helicopter and is modeled by the function f(x) = −15x2 + 6000. The graph below shows the height

melisa1 [442]

General Idea:

Domain of a function means the values of x which will give a DEFINED output for the function.

Applying the concept:

Given that the x represent the time in seconds, f(x) represent the height of food packet.

Time cannot be a negative value, so

$x\geq 0$

The height of the food packet cannot be a negative value, so

$f(x)\geq 0$

We need to replace $-15x^2+6000$ for f(x) in the above inequality to find the domain.

$-15x^2+6000\geq 0 \; \; [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0$

The possible solutions of the above inequality are given by the intervals $(-\infty , -2], [-2,2], [2,\infty )$ . We need to pick test point from each possible solution interval and check whether that test point make the inequality $(x+200)(x-200)\leq 0$ true. Only the test point from the solution interval [-200, 200] make the inequality true.