(c) A. f(x)=3x4−2x3−6x2+6x+1 f′(x)=12x3−6x2−12x+6 f′(x)=6(2x3−x2−2x+1) f′(x)=6(x−1)(2x−1)(x+1) Put, f′(x)=0x=−1,
1
2
,1 f′(x) is increasing in (−1,
1
2
)∪(1,∞) decreasing value in (∞,−1)∪(1∕2,1) minimum value at x=1 or −1 maximum value of x=
1
2
B. f(x)=x+
1
x
f′(x)=1−
1
x2
put f′(x)=0 ∵x2=1⇒x=±1 f′′(x)=
2
x3
Maximum at x=−1 C. f(x)=x4(7−x)3 f′(x)=4x3(7−x)3−3x4(7−x)2 Put f′(x)=0⇒x3(7−x)2(28−7x)=0 x=4 Maximum at x=4 D. f(x)=x4+(8−x)4 f′(x)=4x3−4(8−x)3, put f′(x)=0 x3=(8−x)3⇒x=8−x 2x=8⇒x=4 f′′(x)=12x2+12(8−x)2 f′′(4)≥0,∴ Minimum of x=4