Verified answer

f(x) = 3x^3 + 2x^2 + Cx

f'(x) = 9x^2 + 4x + C

0 = 9x^2 + 4x + C

x = (-4 +/- sqrt(16 - 4 * 9 * C)) / (2 * 9)

We want 16 - 36C to equal 0

16 - 36C = 0

16 = 36C

4 = 9C

4/9 = C

C = 4/9

f(x) = 3x^3 + 2x^2 + cx + 4

f ' (x) = 9x^2 + 4x + c

equate f ' (x) tp zero

9x^2 + 4x + c = 0

=> x^2 + (4/9)x + c/9 = 0

( x + 2/9)^2 - 4/81 + c/9 = 0

in order to have only one critical number, c/9 should be 4/81

c/9 = 4/81

c = 4/9