f(x) = 3x^3 + 2x^2 + cx + 4 f ' (x) = 9x^2 + 4x + c equate f ' (x) tp 0 9x^2 + 4x + c = 0 => x^2 + (4/9)x + c/9 = 0 ( x + 2/9)^2 - 4/80 one + c/9 = 0 with the intention to have in basic terms one extreme quantity, c/9 could be 4/80 one c/9 = 4/80 one c = 4/9

"exactly one critical number" means

"exactly one solution to f ' (x) = 0"

So take the derivative and set it equal to zero

9x² + 4x + C = 0 is a quadratic equation, which has (exactly) one solution iff the discriminant (b² - 4ac) = 0

a = 9

b = 4

c = C

b² - 4ac = 4² - 4(9)(C) = 0

16 - 36C = 0

16 = 36C

C = 4/9

df/dx=9x^2+4x+c=0;

xcr=-2/9+-sqrt(4/81-c/81);

4/81-c/81=0;

c=4/81.