What is the extreme value of y= 3x^2 -3x +5, Is is max. or min?
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calendarEducator since 2008
write3,662 answers
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Given the equation:
y= 3x^2 - 3x + 5
We need to find the extreme value.
First we need to determine the first derivative.
==> y' = 6x -3
Now we will calculate the critical value.
==> 6x -3 = 0 ==> x = 1/2
Now we will determine the values of y(1/2)
==> y(1/2) = 3(1/4) - 3(1/2) + 5 = 3/4 - 3/2 + 5 = ( 3 - 6 + 20)/4 = 17/4
==> y(1/2) = 17/4
Since the coefficient of x^2 is positive, then the parabola is facing up.
Then the parabola has a minimum point at x= 1/2.
Then the extreme values for y is the point ( 1/2, 17/4) and it is a minimum.
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
To find the extreme value of y we need to find the first derivative and equate it to zero to solve for x.
y = 3x^2 - 3x + 15
y' = 6x - 3
equating it to 0
6x - 3 = 0
=> x = 3/6
=> x = 1/2
For x = 1/2, y = 3/4 - 3/2 + 5 = 17/4
The second derivative of y is 6. This is positive for x = 1/2.
Therefore we have a minimum at x = 1/2.
The required extreme value is 17/4, and it is a minimum.
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