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

Question

Charles Arocha · Accepted Answer

#k=20#

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

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

The equation's roots are the critical points:

#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#

Emily Ammerman · Answer

undefined To find the value of $ k $ for which $ f(x) = x^3 - 5x^2 + 3x + k $ has $ 11 $ as its relative minimum, you need to determine the critical points of the function and then use the second derivative test.

1. Find the first derivative of $ f(x) $ and set it equal to zero to find critical points.
   $$ f'(x) = 3x^2 - 10x + 3 $$
   Setting $ f'(x) = 0 $ gives:
   $$ 3x^2 - 10x + 3 = 0 $$
   Solving this quadratic equation will give you the critical points.

2. Once you have the critical points, use the second derivative test to determine whether each critical point corresponds to a relative minimum, maximum, or neither.
   $$ f''(x) = 6x - 10 $$
   Evaluate $ f''(x) $ at each critical point.

3. Since you're looking for the value of $ k $ where the function has a relative minimum of $ 11 $, set $ f(x) = 11 $ and solve for $ x $. Then substitute the found $ x $ into $ f(x) $ to find the corresponding $ k $.