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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 by Veteran (32.9k points)   | 887 views

2 Answers

+4 votes
Best answer

Answer -11

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 (56.6k points)  
selected by
YOU have forgot to check the minimum for end points
not getting @viv
+4 votes

Answer is -11

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 (9.7k points)  
how u have taken the value of x from 1 to 5


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