Question

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

Answer 1

`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`