Prove f(x)=b^x is continuous on (-infinity,infinity) i.e. a exist in the reals. here assume b does not equal 1, but b>0
To prove that the function f(x) = b^x is continuous on the interval (-∞, ∞), we need to show that it is continuous at every point within this interval.
To do this, we can prove that the limit of f(x) as x approaches any real number a exists.
Let's consider an arbitrary point a in the interval (-∞, ∞). We need to find the limit of f(x) as x approaches a.
lim(x→a) b^x = b^a
To prove this, we can use the epsilon-delta definition of limits.
Given any ε > 0, we need to find a value δ > 0 such that if 0 < |x - a| < δ, then |f(x) - f(a)| < ε.
Let's rewrite the expression |f(x) - f(a)|:
|f(x) - f(a)| = |b^x - b^a|
Now, let's factor out b^a from the absolute value:
|b^x - b^a| = |b^a * (b^(x-a) - 1)|
Since b>0 and b≠1, we know that b^a > 0. So, we can simplify the expression:
|b^a * (b^(x-a) - 1)| = b^a * |b^(x-a) - 1|
Now, we need to find a suitable δ that ensures |b^(x-a) - 1| < ε/b^a.
We know that f(x) = b^x is a continuous function for any x, including a. Therefore, b^(x-a) is also continuous for any x. Hence, we can find a δ such that |x - a| < δ implies |b^(x-a) - 1| < ε/b^a.
To find such a δ, we can use logarithms. Taking the natural logarithm (ln) of both sides, we get:
ln(|b^(x-a) - 1|) < ln(ε/b^a)
Now, using the properties of logarithm, we can simplify the inequality:
|x - a| * ln(b) < ln(ε/b^a)
Since ln(b) is a constant and b>0, let's define K = ln(b), which gives us:
|x - a| * K < ln(ε/b^a)
Now, divide both sides of the inequality by K:
|x - a| < ln(ε/b^a) / K
We can define a suitable δ as:
δ = min { ln(ε/b^a) / K }
Now, let's substitute this value of δ back into our proof. If 0 < |x - a| < δ, then:
|x - a| < ln(ε/b^a) / K
By multiplying both sides by K, we get:
|x - a| * K < ln(ε/b^a)
And using the fact that K = ln(b), we have:
|x - a| * ln(b) < ln(ε/b^a)
|x - a| * ln(b) < ln(ε) - ln(b^a)
|x - a| * ln(b) < ln(ε) - a * ln(b)
Now, since |x - a| < ln(ε) / ln(b), we can rewrite the inequality as:
|x - a| * ln(b) < ln(ε) - a * ln(b) < ε
Therefore, if we choose δ = ln(ε) / ln(b), then |x - a| < δ implies |f(x) - f(a)| < ε.
This shows that for any given ε > 0, we can always find a corresponding δ > 0 such that |x - a| < δ implies |f(x) - f(a)| < ε. Hence, the limit of f(x) as x approaches a is f(a) = b^a.
Since this holds true for every point a in the interval (-∞, ∞), we can conclude that the function f(x) = b^x is continuous on (-∞, ∞).