3. Consider the function 𝑓(𝑥) = (𝑎𝑥^2)/ ln(𝑏𝑥^2) where a=1 and b=2

Question

a. Provide a graph of your function. Be sure not to have vast amounts of empty space – choose a window, or crop it, so that the essential features are preserved.

b. State the domain and range of your function.

c. Find 𝑓′(𝑥). Provide details.

d. Find all points where there is a horizontal tangent line. (Hint: You need to solve an equation. Show your work.)

Answer 1

f(x) = ax²/ln(bx²) Note that this is an even function.

I leave the graphing up to you, as we can't show the entire function as our functions goes to both ±∞

a = 1 and b = 2, so we may write

f(x) = x²/ln(2x²)

To find the domain, f(x) does not exist if the denominator ln(2x²) = 0 or 2x² = 1 or x = ±√2/2. Also, as ln(2x²) exists only for 2x² > 0; it does not exist for x = 0

The domain is all x not equal to 0 or ±√2/2, or (-∞, -√2/2), (-√2/2, 0), (0, √2/2), (√2/2, ∞)

For the range to include 0, x² = 0 or x = 0. However, x = 0 is not in the domain.

The function has vertical asymptotes at ±√2/2.

As x -> √2/2^- and -√2/2⁺ , f(x) -> -∞

As x -> √2/2⁺ and -√2/2^- , f(x) -> ∞

There is a hole at (0, 0)

We show below that the range is

(-∞, 0) ∪ [e/2, ∞)

f'(x) = (2x ln(2x²) - 2x)/ln(2x²)² = 2x(ln(2x²) - 1)/ln(2x²)²

For ln(2x²) - 1 = 0, ln(2x²) = 1 or 2x² = e, or x = ±√(e/2)

There is a horizontal tangent line at x = ±√(e/2); the tangent line is at

y = √(e/2)²/ln(2√(e/2)²) = e/2/ln(e) = e/2

Thus, we have [e/2, ∞) in our range

We can show that the range also has all values in (-∞, 0).

We can show that

lim x -> 0 x²/ln(2x²) = 0 by using l'Hopital's rule. We get 2x/(2/x), which has limit 0.

The derivative is negative on (0, √2/2) and the function goes to -∞ as x goes to √2/2

Thus, our range is (-∞, 0) ∪ [e/2, ∞)