Is #f(x)=(x+3)(x-8)(x-3)# increasing or decreasing at #x=1#?

1 Answer
Jun 21, 2016

decreasing

Explanation:

i expect your supposed to follow through on a full product rule etc but there are other ways to look at this.

first compute #f(1) = 4(-7)(-2) = 56#

then compute #delta f = f(1+h) - f(1)# where # h# is a small displacement such that # h \to 0#

#delta f = f(1+h) - f(1) =(4+h)(-7+h)(-2+h) - f(1)#

# = (4+h) (14 -9h +h^2)- f(1)#

the neat thing is we only need look at constant and #h^1# terms, ignoring higher order # h^2, h^3 ,...# terms, because #h \to 0# so those higher order terms are orders of magnitude smaller.

so
#delta f = (4+h) (14 -9h +h^2)- f(1)#

# = 56 + h(-36+14) - 56 + \mathcal{O} (h^2)#

# = -22h + \mathcal{O} (h^2) #

now for # h > 0#, we have #delta f < 0#. for #h < 0#, we have #delta f > 0# so this function is decreasing as it passes through #x = 1#

moreover, the actual slope at #x = 1# should be given by #lim_{h \to 0} [(delta f)/(h) ]_{x=1}= -22 + \mathcal{O} (h) = -22#