Find the Optimum value of the function f(x)=2x^2=6x-10 and also state if the function attains a maximum or minimum.

To find the optimum (maximum or minimum) value of the function f(x) = 2x^2 - 6x - 10, we can use the process of completing the square or differentiate the function and find critical points. Let's use the process of completing the square:

Step 1: Rearrange the function in the standard quadratic form: f(x) = 2x^2 - 6x - 10.
= 2(x^2 - 3x) - 10.

Step 2: Complete the square by adding and subtracting the square of half the coefficient of x. In this case, the coefficient of x is -3, so we need to add and subtract (-3/2)^2 = 9/4.

f(x) = 2(x^2 - 3x + 9/4 - 9/4) - 10.
= 2((x - 3/2)^2 - 9/4) - 10.
= 2(x - 3/2)^2 - 9/2 - 10.
= 2(x - 3/2)^2 - 29/2.

Step 3: Now we can see that the function is in the form f(x) = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola. In this case, the vertex is (3/2, -29/2).

Step 4: Since the leading coefficient (a) is positive (2 in this case), the parabola opens upwards, meaning it attains a minimum value at the vertex.

Therefore, the optimum value of the function f(x) = 2x^2 - 6x - 10 is -29/2, and this minimum value is attained at x = 3/2.

So, the function attains a minimum.