Solution (1)
f(x)=x2−4x+1f′(x)=2x−4f′(2)=2(2)−4=0
Solution (2)
f(x)=x3+2f′(x)=3x2f′(−2)=3(−2)2=12
Solution (3)
f(x)=2x4+3x3−2x+7f′(x)=8x3+9x2−2f′(0)=8(0)3+9(0)2−2=−2
Solution (4)
f(x)=2x+7f′(x)=22x+71(2)=2x+71f′(1)=2(1)+71=91=31
Solution (5)
f(x)=1+x2+3x+6f′(x)=0+2x2+3x+61(2x+3)=2x2+3x+62x+3f′(2)=2(2)2+3(2)+62(2)+3=2167=87
Solution (6)
f(x)=x+4xf′(x)=(x+4)2(x+4)(x)′−(x)(x+4)′f′(x)=(x+4)2x+4−x=(x+4)24f′(−3)=((−3)+4)24=124=4
Solution (7)
f(x)=4−x2x2+3f′(x)=(4−x2)2(4−x2)(2x)−(x2+3)(−2x)f′(x)=(4−x2)28x−2x3+2x3+6x=(4−x2)214xf′(−1)=(4−(−1)2)214(−1)=32−14=−914
Solution (8)
f(x)=x+x3=x+3x−1f′(x)=2x1−3x−2=2x1−x23f′(1)=2(1)1−(1)23=21−3=−25
Solution (9)
f(x)=3−5x−2x9=3−5x−29x−1f′(x)=0−25x1(5)+29x−2=−2x5+2x29f′(2)=−2(2)5+2(2)29=225+89
Solution (10)
f(x)=2x2+3x−1f′(x)=4x+3f′(−1)=4(−1)+3=−1
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