# What is the minimum values of the function v(x) = 3x^2 -3x +5 ?

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To find the minimum value of v(x) = 3x^2 -3x +5, we find the derivative and equate that to zero

v(x) = 3x^2 -3x +5

=> v'(x) = 6x - 3

6x - 3 = 0

=> x = 3/6 = 1/2

For x = 1/2 , v(x) = 3*(1/2)^2 - 3*(1/2) + 5

=> 3*1/4 - 3/2 + 5

=> 3/4 - 3/2 + 5

=> 4.25

Also v''(x) = 6 which is positive for x= 1/2

**Therefore the minimum value of the function is 4.25**

Given the function:

v(x) = 3x^2 - 3x + 5

We need to find the minimum values of v(x).

First we will determine the first derivative v'(x).

==> v'(x) = 6x -3

Now we will calculate the derivatives zero.

==> 6x -3 = 0 ==> 6x = 3 ==> x = 1/2

Then, the critical value for v(x) is the point x = 1/2

Now we will find the values of v(1/2).

==> v(1/2) = 3(1/2)^2 - 3(1/2) +5

= 3/4 - 3/2 + 5 = (3 - 6 + 20)/4 = 17/4

**==> The minimum value of the function v(x) is the point (1/2, 17/4)**

To find the minimum of f(x) = 3x^2-3x+5

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

f(x) 3{x-x+(1/2)^2}+5 - 3(1/2)^2. We added 3(1/2)^2 inside the bracket and subtracted 3(1/2)^2 outside the bracket.

f(x) = 3(x-1/2)^2 +5-3/4

f(x) = 3(x-1/2)^2 +17/4. We see that 3( x-1/2)^2 on the right is always > = 0 and 3(x-1/2)^2 = 0 only when x= 1/2.

So f(x) > = 17/4 and minimum of f(x) = 17/4 when x = 1/2.