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3 years ago

Find the domain of h(x)=x/x^2-x-6

Mathematics

2 answers:

frozen [14]3 years ago

7 0

Set the denominator equal to 0 and solve for x.

x^2 - x - 6 = 0

Factor.

(x - 3)(x + 2 ) = 0

Set each factor to 0 and solve for x.

x - 3 = 0

x = 3

x + 2 = 0

x = -2

The domain is ALL REAL NUMBERS except that x CANNOT equal 3 or -2.

Fynjy0 [20]3 years ago

3 0

Answer:

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Explanation:

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is defined for all Real values of

except those values for which

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So if

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and

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is undefined

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If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

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Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

$2(4x+3)=a(x+7)+b(x-3)$

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

$2(-25)=0-10b$

$-50=-10b$

Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

$2(15)=10a+0$

$30=10a$

Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

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Find the domain of h(x)=x/x^2-x-6​

Find the domain of h(x)=x/x^2-x-6