If # f(x) = ( x^3 )/( x^2 - 25 )#, what are the points of inflection, concavity and critical points?

Question

Evelyn Agard · Accepted Answer

Critical numbers: #0# and #+-sqrt75# Concave down on #(-oo,-5)# and on #(0,5)# Concave up on #(-5,0)# and on #(5,oo)# Inflection point #(0,0)# #f'(x) = (x^2(x^2-75))/(x^2-25)^2# #f'(x)# DNE at #x=+-5# but those are not in the domain, so they are not critical. #f'(x) = 0# at #x=0# and at #x + +-sqrt75# which are in the domain, so they are all critical numbers. #f''(x) = (50x(x^2+75))/(x^2-25)^3# could change sign at #0# and at #+-5# On #(-oo,-5)#, #f''(x) < 0#, so #f# is concave down. On #(-5,0)#, #f''(x) > 0#, so #f# is concave up. On #(0,5)#, #f''(x) < 0#, so #f# is concave down. On #(5,oo)#, #f''(x) < 0#, so #f# is concave up. The concavity changes at #x=-5#, #0# and #5#. An inflection point is a point of the graph where concavity changes. Since #-5# and #5# are not in the domain of #f#, there are no IPs there, but #f(0) = 0#, so #(0,0)# is an Infle pt.

Ezra Adams · Answer

undefined To find the points of inflection, concavity, and critical points, we first need to find the second derivative of $ f(x) $, then set it equal to zero to find critical points, and finally test for concavity using the second derivative test. First derivative: $ f'(x) = \frac{d}{dx}\left(\frac{x^3}{x^2 - 25}\right) $ Using the quotient rule: $ f'(x) = \frac{(x^2 - 25)\cdot 3x^2 - x^3\cdot 2x}{(x^2 - 25)^2} $ Simplifying: $ f'(x) = \frac{3x^4 - 75x^2 - 2x^4}{(x^2 - 25)^2} $ Further simplification: $ f'(x) = \frac{x^4 - 75x^2}{(x^2 - 25)^2} $ Second derivative: $ f''(x) = \frac{d}{dx}\left(\frac{x^4 - 75x^2}{(x^2 - 25)^2}\right) $ Using the quotient rule and simplifying: $ f''(x) = \frac{2x(x^4 - 75x^2)(2x^2 - 50) - (x^4 - 75x^2)(2x)}{(x^2 - 25)^3} $ Further simplification: $ f''(x) = \frac{2x(x^5 - 50x^3 - 75x^2) - 2x(x^5 - 50x^3 - 75x^2)}{(x^2 - 25)^3} $ $ f''(x) = \frac{4x^6 - 200x^4 - 150x^3 - 4x^6 + 200x^4 + 150x^3}{(x^2 - 25)^3} $ $ f''(x) = 0 $ for all $ x $ Critical points: Since the second derivative is always zero, there are no critical points. To find the points of inflection, we look for the values of $ x $ where the concavity changes. This happens when the second derivative is zero or undefined, or at vertical asymptotes. Since the second derivative is always zero, there are no points of inflection. Concavity: Since the second derivative is always zero, the concavity cannot be determined using the second derivative test. We can analyze the behavior of $ f(x) $ near its vertical asymptotes to determine concavity. Asymptotes: Vertical asymptotes occur when the denominator of $ f(x) $ equals zero. Therefore, $ x^2 - 25 = 0 $ gives $ x = ±5 $. These are vertical asymptotes. Concavity Analysis: - For $ x < -5 $, $ f''(x) > 0 $, so $ f(x) $ is concave up. - For $ -5 < x < 5 $, $ f''(x) < 0 $, so $ f(x) $ is concave down. - For $ x > 5 $, $ f''(x) > 0 $, so $ f(x) $ is concave up. Therefore, the function $ f(x) $ is concave up for $ x < -5 $ and $ x > 5 $, and concave down for $ -5 < x < 5 $.