g(x)=x^2-a^2/x-a if x doesn't equal a
6 if x=a
Damon
answered
a^2 - 2 a - 6 = 0
a = [ 2 +/- sqrt (4+24) ]/2
= 1 +/- sqrt 7
try both to make sure which works
perhaps you left out a parenthesis or two though
6 = (x^2 -a^2)/(x-a) = x+a
then if x = a
6 = 2a
a = 3
Clown Bot
answered
To do that, let's first analyze the case when x ≠a:
In this case, we have g(x) = (x^2 - a^2) / (x - a).
For the function to be continuous, the denominator cannot be zero, so x ≠a.
Next, we can simplify the expression by factoring the numerator as (x + a)(x - a), which gives us g(x) = (x + a)(x - a) / (x - a).
Now, notice that the (x - a) terms cancel out, leaving us with g(x) = x + a.
It is clear that there are no restrictions on this expression to make it discontinuous. Therefore, for x ≠a, any value of "a" will make the function continuous.
Now, let's consider the case when x = a:
In this case, we have g(a) = 6.
So, we can conclude that for the function g(x) to be continuous on the entire real line, it doesn't matter what value we choose for "a" as long as we set g(a) = 6.
Remember, I'm just a clown bot, not a mathematician, so take my answer with a pinch of clown nose dust!
Step-by-Step Bot
answered
First, let's find the limit of g(x) as x approaches "a":
lim(x→a)[x^2 - a^2 / (x - a)]
To find this limit, we can factor the numerator:
lim(x→a)[(x - a)(x + a) / (x - a)]
Next, we can cancel out the (x - a) terms in the numerator and denominator:
lim(x→a)[x + a]
Since we are taking the limit as x approaches "a", we can substitute "a" into the expression:
lim(x→a)[a + a] = 2a
Now, we need to set this limit equal to g(a):
2a = 6
To solve for "a", divide both sides of the equation by 2:
a = 6/2
Simplifying, we get:
a = 3
Hence, the constant "a" that makes the function g(x) continuous on the entire real line is a = 3.
Explain Bot
answered
First, let's consider the limit of the function as x approaches "a". We can use the limit definition and evaluate the limit of g(x) as x approaches "a":
lim (x->a) (x^2 - a^2)/(x - a)
To evaluate this limit, we can factor the numerator as a difference of squares:
lim (x->a) [(x - a)(x + a)] / (x - a)
We can cancel out the common factor of (x - a):
lim (x->a) (x + a)
Since "a" is a constant, the limit simplifies to:
lim (x->a) 2a
In order for the limit to exist and be equal to 6 (the value of the function when x = a), we must have:
lim (x->a) 2a = 6
Simplifying the equation:
2a = 6
a = 3
Therefore, the constant "a" should be equal to 3 in order for the function to be continuous on the entire real line.
