Question #9822d

1 Answer
Jim S
Dec 13, 2017

#m=6# , #k=8#

Explanation:

Too late but i found a solution!

#f(x)=2x^2+kx+m , x##inRR# ,
#kin##RR#, #m##in##RR#

#f# continuous and differentiable in #RR# with
#f'(x)=4x+k#

  • #f(-2)=-2 <=> -2=2(-2)^2+k(-2)+m#

#<=># #-2=8-2k+m#

#<=># #m=2k-10# #(1)#

#f# is defined in #RR#
#f# is differentiable in #RR# and so #-2inRR#
#f# has local extrema at #-2#

-- According to #color(red)(Fermat)# Theorem regarding

stationary points, the derivative of #f# at #-2# must equal #0#

In other words , #f'(-2)=0# #<=># #4(-2)+k=0# #<=>#

#<=># #k=8# #(2)#

So #(1)# now gives #m=2*8-10=6#

So #m=6#, #k=8# and #f# now becomes #f(x)=2x^2+8x+6#

#=# #2(x^2+4x+3)# #=# #2(x+1)(x+3)#

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