Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
Move    the right-hand  term    to  the other   side    of  the equals  sign:   

Cross-multiply: 2(100   −   x^2 )   =   2x^2

Distribute: 200 −   2x^2    =   2x^2

And solve:  200 =   4x^2

x   =       =   5 , so  2x  =   10

Now we  can solve   for y:  y   =   

Therefore,  the dimensions  are 10  by  10.


  1. Maximum at (0, 0); Minima at (2, −4) and (−2, −4); Points of inflection at and


.

First,  we  can easily  see that    the graph   has x-intercepts    at  x   =   0   and x   =   ± . Next,   we  take    the

derivative:     =   x^2     −4x.    Next,   we  set the derivative  equal   to  zero    to  find    the critical    points.

There   are three   solutions:  x   =   0,  x   =   2,  and x   =   −2.

We  plug    these   values  into    the original    equation    to  find    the y-coordinates   of  the critical    points:

When    x   =   0,  y   =       −2(0)^2     =   0.  When    x   =   2,  y   =       −2(2)^2     =   −4. When    x   =   −2, y   =   

−2(−2)^2    =   −4. Thus,   we  have    critical    points  at  (0, 0), (2, −4),    and (−2,    −4).    Next,   we  take    the

second  derivative  to  find    any points  of  inflection. The second  derivative  is      =   3x^2    −   4,

which   is  equal   to  zero    at  x   =       and x   =   − . We  plug    these   values  into    the original    equation

to  find    the y-coordinates.  When    x   =       :    .  When    x   =   −
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