The inverse of a function is a function. always /sometimes /never

Mathematics

2 answers:

Liono4ka [1.6K]3 years ago

7 0

The answer is sometimes

Lisa [10]3 years ago

4 0

Hmmm,
a function can be determined by the vertical line test
if you can draw a vertical line that passes through the graph twice, it is not a function

the inverse is the graph reflected across the line y=x
we can determine the answer to this question graphically

take y=x^2 for example
this is a function
reflected across the line y=x, and it is not a function
it fails the vertical line test

but, we can also take y=3x and reflect it and it is still a function

so the answer is sometimes

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$2,500 at 6% for 3 years

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3 years ago

Factorize of 2a³-a²+a-2

IrinaK [193]

<h3>Answer: (a - 1)(2a² + a + 2)</h3>

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Explanation:

Use the rational root theorem to determine this list of possible rational roots: 1, -1, 1/2, -1/2

Plug each possible root one at a time into the original expression given. If the simplified result is 0, then that possible root is an actual root.

If we tried say a = -1, then,

2a³-a²+a-2 = 2(-1)³-(-1)²+(-1)-2 = -6

The result is not zero, so a = -1 is not an actual root.

But if we tried say a = 1, then,

2a³-a²+a-2 = 2(1)³-1²+1-2 = 0

We get 0 so a = 1 is an actual root. I'll let you try the other values, but you should find that a = 1 is the only rational root.

Since a = 1 is a root, this makes (a-1) to be a factor.

From here, use either synthetic or polynomial long division to determine the other factor. Refer to the diagram below for each method.

Regardless of which method you pick, the quotient is 2a² + a + 2 which is the other factor needed. The remainder of 0 tells us we have (a-1) as a factor. For more information, check out the remainder theorem.