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Virty [35]
3 years ago
12

The range of the function y=secx-2 is all reals except -1 1 -3 -2

Mathematics
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

For a function y = f(x), the range is the set of all the possible values of y.

In the question you wrote:

y = secx - 2

This can be interpreted as:

y = sec(x - 2)

or

y = sec(x) - 2

So let's see each case (these are kinda the same)

If the function is:

y = sec(x - 2)

Firs remember that:

sec(x) = 1/cos(x)

then we can rewrite:

y = 1/cos(x - 2)

notice that the function cos(x) has the range -1 ≤ y ≤ 1

Then for the two extremes we have:

y = 1/1 = 1

y = 1/-1 = -1

Notice that for:

y = 1/cos(x - 2)

y can never be in the range  -1 < x < 1

As the denominator cant be larger, in absolute value, than 1.

Then we can conclude that the range is all reals except the interval:

-1 < y < 1

If instead the function was:

y = sec(x) - 2

y = 1/cos(x) - 2

Then with the same reasoning, the range will be the set of all real values except:

-1 - 2 < y < 1 - 2

-3 < y < -1

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Explanation:

  • The numerator has three terms while the denominator only has one term. We divide the numerator terms each separately with the denominator term.
  • So 8x³ + 6x² - 4x / -2x becomes (8x³ / -2x) + (6x² / -2x) + (- 4x/ -2x).
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  • Adding all the terms we get -4x² - 3x + 2. This is the option a.
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