Final answer:
To find the value of > 0 such that |f(x) − L| < 0.01 whenever 0 < |x − 4| < , we need to determine the value of f(x) when |x - 4| < . Given that f(x) = x^2 + 3, we substitute x = 4 into the function to find f(4) = 19. Therefore, for |x - 4| < , L = 19. Now we can set up the inequality |f(x) − L| < 0.01 and solve for .
Step-by-step explanation:
To find the > 0 such that |f(x) − L| < 0.01 whenever 0 < |x − 4| &, we need to first determine the value of f(x) when |x - 4| < .
Given that f(x) = x^2 + 3, we substitute x = 4 into the function to find f(4) = (4^2) + 3 = 16 + 3 = 19.
Therefore, for |x - 4| < , L = 19. Now we can set up the inequality |f(x) − L| < 0.01 and solve for .
|f(x) − L| = |(x^2 + 3) - 19| = |x^2 - 16| < 0.01
Since we know that |x - 4| < , we also know that when 0 < |x - 4| < , |x - 4| = 4 - x.
This means our inequality becomes |(4 - x)(4 + x)| < 0.01.
To simplify further, we can apply the difference of squares: |16 - x^2| < 0.01.
We find that -0.01 < 16 - x^2 < 0.01.
By rearranging the equation, we have -0.01 + x^2 < 16 and x^2 < 16.01, which means -4.001 < x < 4.001.
Therefore, the answer is |x - 4| < 4.001.