How do you find the value of k for which #f(x)=x^3-5x^2+3x+k# has 11 as its relative minimum?

1 Answer
Jan 12, 2017

#k=20#

Explanation:

First we find where #f(x)# has its local extrema:

#f'(x) = 3x^2-10x+3#

The critical points are roots of the equation:

#3x^2-10x+3 = 0#

#x = frac (5+- sqrt (25-9)) 3 = (5+- 4)/3#

#x_1 = 1/3#, #x_2 = 3#

As #f'(x)# is a second order polynomial with positive leading coefficient, it has positive values in the intervals outside the roots, and negative between the roots, that is:

#f'(x) >0 # for #x in (-oo,1/3) uu (3,+oo)#

#f'(x) <0 # for #x in (1/3,3) #

which means that #x_1# is a local maximum and #x_2# the only local minimum.

Now we can find #k# by posing #f(x_2) = 11#

#x_2^3-5x_2^2+3x_2+k = 11#

#27-45+9+k =11#

#k=20#