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3 years ago

Function Operations Picture attached

Mathematics

1 answer:

sergeinik [125]3 years ago

6 0

Answer: Second option

Step-by-step explanation:

You need to make the multiplication of the function $c(x)=\frac{5}{x-2}$ and the function $d(x)=x+3$ . Then:

$(cd)(x)=(\frac{5}{x-2})(x+3)$

You need to apply the Distributive property:

$(cd)(x)=\frac{5(x+3)}{x-2}\\\\(cd)(x)=\frac{5x+15}{x-2}$

Therefore, the domain will be all the number that make the denominator equal to zero.

Then, make the denominator equal to zero and solve for x:

$x-2=0\\x=2$

Therefore, the domain is: All real values of x except $x=2$

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I've done every problem I only need with the help this problem

bulgar [2K]

Answers:

$\displaystyle \lim_{x \to 2^{+}} f(x) = 1\\\\\displaystyle \lim_{x \to 2^{-}} f(x) = 1\\$

Both result in the same limit value. This allows us to say $\displaystyle \lim_{x \to 2} f(x) = 1$ without the plus or minus over the 2.

The left and right hand limits may not always match like this.

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Explanation:

The notation $\displaystyle \lim_{x \to 2^{+}} f(x)$ means that we are approaching x = 2 from the right hand side. This is from the positive direction. So we start at say x = 3 and move to x = 2.5 then to x = 2.1 then to x = 2.01 and so on.

Because we started with values x > 2, we will use the third definition of the piecewise function

if x > 2, then f(x) = 3x-5

Plug in x = 2 to get

f(x) = 3x-5

f(2) = 3(2)-5

f(2) = 6-5

f(2) = 1

This shows $\displaystyle \lim_{x \to 2^{+}} f(x) = 1$

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For the other limit, we're approaching x = 2 from the negative side. So we could start at say x = 0, then move to x = 1, then to x = 1.5 then to x = 1.9 then to x = 1.99, and so on.

We're using x values such that x < 2 now.

So we'll be using the first definition of the piecewise function

If x < 2, then f(x) = x^2 - 3

f(x) = x^2-3

f(2) = 2^2-3

f(2) = 4-3

f(2) = 1

We end up with $\displaystyle \lim_{x \to 2^{-}} f(x) = 1$

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Both right hand limit and left hand limit result in the same value

Because $\displaystyle \lim_{x \to 2^{+}} f(x) = \displaystyle \lim_{x \to 2^{-}} f(x) = 1$

We can shorten that to $\displaystyle \lim_{x \to 2^{}} f(x) = 1$ meaning we can approach x = 2 from either direction to arrive at the same limiting value.