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Easy

Question

The smallest value of x2 – 3x + 3 in the interval open square brackets negative 3 comma fraction numerator 3 over denominator 2 end fraction close square bracketsis

  1. 1    
  2. 5    
  3. – 15    
  4. ¾    

hintHint:

We are given an function. We have to find the smallest value of the function in the given interval. It means we will find a point at which the given value is minimum. Then we will substitute the value and find the minimum value of the function.

The correct answer is: ¾


    The given function is  x2 - 3x + 3.
    The given interval is open square brackets negative 3 space 3 over 2 close square brackets
    f(x) = x2 - 3x + 3
    We will take the first derivative of the function and equate it to zero.
    f'(x) = 2x - 3
    0 = 2x - 3
    Adding 3 to both the sides
    3 = 2x
    Rearranging the equation and dividing both the sides by 2
    x = 3 over 2
    This is the value of minima. This is the point where the given function gas minimum value. This point is included in the interval.
    So the value of the function at this point will be minimum value.
    f left parenthesis x right parenthesis space equals space x squared space minus space 3 x space plus space 3
f left parenthesis 3 over 2 right parenthesis space equals left parenthesis 3 over 2 right parenthesis squared space minus 3 left parenthesis 3 over 2 right parenthesis space plus space 3
space space space space space space space space space space space equals space 9 over 4 minus space 9 over 2 plus 3
space space space space space space space space space space space space equals fraction numerator 9 space minus space 18 space plus space 12 over denominator 4 end fraction
space space space space space space space space space space space space equals 3 over 4
S o comma space t h e space m i n i m u m space v a l u e space o f space g i v e n space f u n c t i o n space i s space 3 over 4

    The alternate way to solve the question is to put both the values of interval of the function. Then we compare the values. We also check if the first derivative is postive or negative. If it's negative, the given function is decreasing. And if postive, the given function is increasing.

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