GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 58 of 65

Marks: +1, -0

Let

f(x)=3 x^{3}-7 x^{2}+5 x+6

. The maximum value of

f(x)

over the interval

[0,2]

is__________ (upto 1 decimal place).

Solution:

f(x)=3 x^{3}-7 x^{2}+5 x+6

f^{'}(x)=9 x^{2}-14 x+5

f^{' '}(x)=18 x-14

f^{'}(x)=0

x^{2}-14 x+5=0

x=1,0.55

x=1

f^{' '}(1)=18-14=4>0

, minima

x=0.55

f^{' '}(0.55)=-4.1<0

maxima
Maximum

\{f(0), f(0.55), f(2)\}

Maximum

\{6,7.13,12\}=12

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