How do you find intercepts, extrema, points of inflections, asymptotes and graph #f(x)=(4x)/(sqrt(x^2+15))#?

Question

Scarlett Allison · Accepted Answer

No intercepts. Horizontal asymptotes : #rarr y = 4 and y =-4 larr#
Point of inflexion : O ( see the graph ),
No extrema, as #y in (-oo, oo)#.

Use #y = f(x) = ( 4 x )/(|x| sqrt(1+15/x^2)# If (x, y) is on the graph, so is (-x, -y), showing symmetry about O. As #x to +-oo, y to +-4# Cross multiply and differentiate to get the following results. y' simplifies to #60/(x^2+15)^1.5 in (0, 4sqrt15]# y, y and y''. vanish at O and #y''' ne 0#, revealing O as the POI.. graph{(4x/sqrt(x^2+15)-y)(y+4)(y-4)=0 [-10, 10, -5, 5]}

Luke Alvarez · Answer

undefined To find intercepts, set $ f(x) = 0 $ for x-intercepts and $ x = 0 $ for the y-intercept.

1. X-intercepts: Set $ f(x) = 0 $ and solve for x.
   $ \frac{4x}{\sqrt{x^2+15}} = 0 $
   $ 4x = 0 $
   $ x = 0 $
   So, the x-intercept is at (0, 0).

2. Y-intercept: Substitute $ x = 0 $ into the function.
   $ f(0) = \frac{4(0)}{\sqrt{0^2+15}} $
   $ f(0) = 0 $
   So, the y-intercept is at (0, 0).

To find extrema, points of inflection, and asymptotes, you'll need to differentiate the function.

3. Differentiate $ f(x) $ with respect to $ x $ to find critical points.
   $ f'(x) = \frac{d}{dx}\left(\frac{4x}{\sqrt{x^2+15}}\right) $
   Use the quotient rule:
   $ f'(x) = \frac{4\sqrt{x^2+15} - 4x \cdot \frac{1}{2}(x^2+15)^{-\frac{1}{2}} \cdot 2x}{x^2+15} $
   Simplify and set $ f'(x) = 0 $ to find critical points.

4. To find points of inflection, find the second derivative $ f''(x) $ and determine where it changes sign.

5. To find asymptotes, analyze the behavior of the function as $ x $ approaches positive or negative infinity.

For the given function $ f(x) = \frac{4x}{\sqrt{x^2+15}} $:

- To find extrema and points of inflection, differentiate and find critical points.
- To find asymptotes, analyze the behavior of the function as $ x $ approaches positive or negative infinity.