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What is the least value of the function $f(x) = 2x^{2}-8x-3$ in the interval $[0, 5]$?

1. $-15$
2. $7$
3. $-11$
4. $-3$
asked in Calculus | 852 views

f(x) = 2x2 -8x -3

f'(x)=4x-8

f''(x)=4 , then it has minimal value at the point x=2

minimum value is 2⨉(2)2 -8⨉(2) -3= -11

answered by Veteran (54.8k points)
selected
YOU have forgot to check the minimum for end points
not getting @viv

Substutue all the value of x from 1 to 5

in case of 0 it is -3

in case of 1 you will get -9

in case of 2 you will get -11

in case of 3 you will get -9

incase of 4 you will get -3

incase of 5 you will get 7

so the least value is -11 -->option c

answered by Boss (9.6k points)
how u have taken the value of x from 1 to 5