limit as x approaches 3 where f(x)=6x+2
Find a formula for δ such that for any ε > 0, |f (x) - 20| < ε if |x - 3| < δ. (In your answer, use the word "epsilon" instead of the symbol ε.)
I have no idea how to do this? Please help!!!
To find a formula for δ, we need to work backwards from the given condition |f(x) - 20| < ε if |x - 3| < δ. Here's how you can approach it step by step:
1. Start with the condition |f(x) - 20| < ε, where f(x) = 6x + 2, and replace f(x) with its expression:
|6x + 2 - 20| < ε
2. Simplify the expression inside the absolute value:
|6x - 18| < ε
3. Factor out 6 from the expression:
6 |x - 3| < ε
4. Divide both sides by 6:
|x - 3| < ε/6
So, we have |x - 3| < ε/6. This suggests that if we choose δ to be ε/6, the condition will hold.
Therefore, the formula for δ is δ = ε/6.
To summarize, for any ε > 0, if |x - 3| < δ = ε/6, then the condition |f(x) - 20| < ε will be satisfied.