Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
piecewise.  Thus,   s(5)    −   s(2)    =   568 −   49  =   519.


  1. Let f be the function given by f(x) = −2x^4 + 6x^2 + 2.


(a) Find    the equation    for the line    normal  to  the graph   at  (1, 6).

The line    normal  to  the graph   will    have    a   slope   that    is  the opposite    reciprocal  of  the tangent line

at  that    point.  Therefore,  begin   by  finding the slope   of  the tangent line,   i.e.,   the first   derivative. f′

(x) =   −8x^3   +   12x and f′  (1) 4.  The slope   of  the normal  line    is  − . So, the equation    of  the normal

line    is: y   −   6   =   − (x    −   1).

(b) Find    the x-  and y-coordinates   of  the relative    maximum and minimum points.

The relative    maximum and minimum will    occur   at  the points  when    the first   derivative  is  zero    or

undefined.  In  this    case,   set the first   derivative  to  zero    and solve   for x:  f′(x)   =   −8x^3   +   12x =   0,

and x   =   0,  x   =    ,  and x   =   −    .  To  determine   which   of  these   is  a   relative    maximum and

which   is  a   relative    minimum,    find    the second  derivative  at  each    of  these   critical    points. f′′ (x) =

−   24x^2 +  12,     and    f′′  (0)     =   12,    f′′  =   −   24,     and    f′′  =   −24.    By  the     second

derivative  test,   x   =   0   corresponds with    a   relative    minimum because f′′(0)  >   0   and x   =    ,  and

x   =   −       correspond  with    relative    maximums    because f′′ and f′′ <    0.  To

determine   the y-coordinates   of  these   points, plug    them    back    into    f(x):   f   (0) =   2,  f   =   

and f   =    .  So, a   relative    minimum occurs  at  (0, 2)  and relative    maximums    occur   at  

    and .

(c) Find    the x-  and y-coordinates   of  the points  of  inflection.

Points  of  inflection  occur   when    the second  derivative  equals  zero.   Take    the second  derivative
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