Tentukan nilai dan titik stasioner fungsi f (x)=x³-6x²+9x+1​

Jawab:

turunan
titik stationer, jika  f'(x) = 0

f(x) =x³-6x²+9x+1​
f'(x)= 0
3x² -12x  +  9= 0
3(x² – 4x + 3) =0
3(x – 1)(x – 3) =0
x=  1  atau  x = 3

nilai statiober

f(x) =x³-6x²+9x+1​
f(1) = (1)³- 6 (1)²+ 9(1) + 1
f(1) = 1 – 6 + 9  +1
f(1) = 5

f(3) = (3)³ – 6 (3)²+ 9(3) + 1
f(3) = 27-  54 +  27 + 1
f(3) = 1

nilai  stationer f(x)=  1 dan 5
titik  stationer (x, f(x)} = (1, 5)  dan (3, 1)​