calulus help plz

Question

) Find all the real values of a and b such that the function f(x)=\ [[(sin^2 (2x))/(a * x ^ 2), ifx > 0], [b, ifx = 0], [arctan(1/(x ^ 2)), ifx < 0]] is continuous.

Answer 1

Hey Josh! Let's break this down: we have a function split into three parts at x = 0.

To the right of x = 0, the function is (sin²(2x))/(ax²).

At x = 0, the function value is "b".

To the left of x = 0, the function is arctan(1/x²).

For the function to be continuous, all three parts must equal the same value at x = 0. In other words, the limit as x approaches zero from both sides must equal the function value at zero.

For the part less than zero, the limit as x approaches zero of arctan(1/x²) equals π/2.

Here's why: As x approaches infinity, 1/x² approaches zero. The value for x where tan(x) approaches infinity is π/2. And since tan(x) and arctan(x) are inverses, we get π/2.

Since π/2 must equal the function value at zero, then b = π/2.

Now, the section to the right of zero must also equal π/2. So:

(sin²(2x))/(ax²) = π/2

We solve for "a":

sin²(2x) = (π/2)(ax²)

2sin²(2x) = πax²

(2sin²(2x))/(x²) = πa

(2(sin(2x))/x) * (sin(2x)/x) = πa

Let 2x = w, meaning x = w/2. Substituting:

(2(sin(w))/(w/2)) * (sin(w)/(w/2)) = πa

4(sin(w))/w * 2(sin(w))/w = πa

Since the limit as w approaches zero of (sin(w))/w = 1, then:

4(1)(2)(1) = πa

8 = πa

So, a = 8/π.

Answer 2

This piecewise function has a boundary at x = 0 where three different things occur. To the right of x = 0, the function value is found by (sin²(2x))/(ax²). At x = 0 the function value is "b" and to the left of x = 0, the function value is found by arctan(1/x²). To be continuous, all three pieces MUST have the same function value at zero. (the correct terminology would be the the limit as x approaches zero from the left must equal f(0) which must equal the limit as x approaches zero from the right).

For the section less than zero, the limit as x approaches zero of arctan(1/x²) = π/2

Rationale: As x approaches infinity, then 1/x² approaches zero. The value for x where tan(x) approaches infinity is π/2 and since tan(x) and arctan(x) are inverses then we get π/2.

And since π/2 MUST equal the function value at zero, then b = π/2

Since the section to the right of zero must also equal π/2 then:

(sin²(2x))/(ax²) = π/2 and solve for "a":

sin²(2x) = (π/2)(ax²)

2sin²(2x) = πax²

2sin²(2x)/(x²) = πa

2(sin(2x))/x•(sin(2x))/x = πa

Let 2x = w meaning x = w/2. Substituting:

2(sin(w))/(w/2)•(sin(w))/(w/2) = πa

4(sin(w))/w • 2(sin(w))/w = πa

since the limit as w approaches zero of (sin(w))/w = 1 then:

4(1)(2)(1) = πa

8 = πa

a = 8/π