How do you find intercepts, extrema, points of inflections, asymptotes and graph #g(x)=x+32/x^2#?

Question

Isabella Ackerman · Accepted Answer

Please see below for a partial solution.

#g(x) = x+32/x^2# x intercepts #g(x) = 0# at #x=-32/x^2# which happens at #x^3 = -32# so #x = root(3)(-32) = -2root(3)4# y intercept None. #g(0)# does not exist. Asymptotes #lim_(xrarr0)g(x) = oo#, so #x=0# (the #y#-axis) is a verticle asymptote.. #lim_(xrarr00)g(x) = oo# so there is no horizontal asymptote. #lim_(xrarroo)(g(x)-x) = 0# so #y=x# is an oblique (slant) asymptote) Analysis of first derivative #g'(x) = 1-64/x^3 = (x^3-64)/x^3# is undefined at #x=0# and is #0# at #x=4#. On #(-oo,0)#, we have #g'(x) > 0# so #g# is increasing. #x=0# is not a critical number. On #(0,4)#, we have #g'(x) < 0# so #g# is decreasing. On #(4,oo)#, we have #g'(x) > 0# so #g# is increasing. #f(4)=6# is a local minimum.

Isaiah Allen · Answer

undefined To find the intercepts:
- x-intercept: Set $ g(x) = 0 $ and solve for $ x $.
- y-intercept: Evaluate $ g(x) $ when $ x = 0 $.

To find extrema:
- Take the derivative of $ g(x) $, $ g'(x) $.
- Set $ g'(x) = 0 $ and solve for $ x $ to find critical points.
- Test these critical points to determine if they correspond to local extrema.

To find points of inflection:
- Take the second derivative of $ g(x) $, $ g''(x) $.
- Set $ g''(x) = 0 $ and solve for $ x $ to find possible inflection points.
- Test these points to confirm if they are points of inflection.

To find asymptotes:
- Horizontal asymptote: As $ x $ approaches positive or negative infinity, find the limit of $ g(x) $.
- Vertical asymptotes: Find values of $ x $ where the denominator of $ g(x) $ equals zero.

Graph $ g(x) = x + \frac{32}{x^2} $ using the information gathered from intercepts, extrema, points of inflection, and asymptotes.