f(x)= x^3/x^2-16 defined on [-19, 16]
How would you find the vertical asymptopes? and the inflection point?
I think you take the derivative and set it to zero? but i cant figure out the correct derivative or something??
assume you mean (x^2-16) on the bottom
vertical asymptotes are where the denominator goes to zero
that is when x = -4 and x = +4
derivative of fraction =
(bottom * derivative of top - top * derivative of bottom)
all over bottom squared
[(x^2-16)*3 x^2 - x^3(2x) ]/bottom^2
look for zeros of numerator
3 x^4 - 48 x^2 - 2 x^4
=x^4 - 48 x^2
= x^2 (x^2-48)
= x^2 (x-sqrt 48)(x+sqrt 48)
so zero at -sqrt 48, 0 and + sqrt 48
check my arithmetic, I did that fast.
by the way sqrt 48 = 4 sqrt 3
To find the vertical asymptotes, we need to check for any values of x that make the denominator of the fraction in the given function zero. In this case, the denominator is x^2 - 16.
Setting the denominator equal to zero, we get x^2 - 16 = 0. We can factor this equation as (x + 4)(x - 4) = 0, which gives us two possible values for x: x = -4 and x = 4. These are the vertical asymptotes.
To find the inflection point, we need to find the second derivative of the function and set it equal to zero. Let's start by finding the first derivative of f(x).
Using the quotient rule, the first derivative of f(x) = x^3/(x^2 - 16) is:
f'(x) = [(3x^2)(x^2 - 16) - (x^3)(2x)] / (x^2 - 16)^2.
Now, simplify this expression:
f'(x) = (3x^4 - 48x^2 - 2x^4) / (x^2 - 16)^2.
f'(x) = (x^4 - 48x^2) / (x^2 - 16)^2.
To find the second derivative, we differentiate f'(x) with respect to x:
f''(x) = [(4x^3)(x^2 - 16)^2 - 2(x^4 - 48x^2)(2x)(x^2 - 16)] / (x^2 - 16)^4.
Simplify this expression:
f''(x) = (4x^5 - 64x^3 - 4x^5 + 192x^3) / (x^2 - 16)^3.
f''(x) = (128x^3) / (x^2 - 16)^3.
Now, we set f''(x) equal to zero and solve for x:
(128x^3) / (x^2 - 16)^3 = 0.
Since the numerator is nonzero, there are no solutions to this equation. Therefore, there are no inflection points in the given function f(x) = x^3/(x^2 - 16) on the interval [-19, 16].