If f(x) = x2 – 25 and g(x) = x – 5, what is the domain of (f/g)(x)?

Mathematics

2 answers:

3241004551 [841]3 years ago

7 0

Answer:

b) all real values of x except x = 5

Step-by-step explanation:

Given functions,

$f(x) = x^2 - 25$

$g(x) = x-5$

$(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$

By substituting the values,

$(\frac{f}{g})(x)=\frac{x^2-25}{x-5}$

Which is a rational function,

Since, a rational function is defined for all real values except those for which the denominator = 0,

That is,

x - 5 = 0

⇒ x = 5

Hence, the domain of (f/g)(x) are all real values of x except x = 5 ,

Option b) is correct.

Stels [109]3 years ago

5 0

Choice b. is right because for x=5 will result this fraction undefined bc than for x=5 the denominator will be equal zero

You might be interested in

Jake sold 36 tickets to the school fair and jeanne sold 15 what is the ratio

artcher [175]

12:5 is what you are looking for. 12/5 in fraction form

3 0

3 years ago

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.