We have f(x)=x3−4x2+4x+3 ∴‌‌f′(x)=3x2−8x+4 Now, f′(x)=0⇒3x2−8x+4=0 ⇒‌‌3x2−6x−2x+4=0 ⇒‌‌3x(x−2)−2(x−2)=0 ⇒‌‌(x−2)(3x−2)=0⇒x=2,‌
2
3
Now, f(−1)=−1−4−4+3=−6 f(2/3)=‌
8
27
−4×‌
4
9
+4×‌
2
3
+3 =‌
8−48+72+81
27
=‌
113
27
f(2)=8−14+8+3=5 f(3)=27−36+12+3=6 ∴f(x) has minimum value at x=−1