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 value of the function f(x) = 2x^2 + 6x - 10 and determine if it attains a maximum or minimum, we can use the concept of calculus.

Step 1: Find the derivative of the function f(x) with respect to x.
The derivative of f(x) can be found by applying the power rule of differentiation. For each term, multiply the coefficient by the exponent and reduce the exponent by 1. In this case, the derivative of f(x) = 2x^2 + 6x - 10 will be:

f'(x) = 4x + 6

Step 2: Set the derivative f'(x) equal to zero and solve for the value of x.
To find the stationary points (points where the derivative is zero), we set f'(x) equal to zero and solve for x:

4x + 6 = 0

Subtract 6 from both sides:
4x = -6

Divide both sides by 4:
x = -6/4
x = -3/2 or -1.5 (rounded to decimal places)

Step 3: Determine if it's a maximum or minimum.
To determine if the function attains a maximum or minimum at x = -3/2, we need to examine the second derivative of f(x). The second derivative gives us information about the concavity of the function.

Taking the derivative of f'(x), we get the second derivative:

f''(x) = 4

Since the second derivative is positive (4 > 0), it indicates that the function is concave up everywhere, and at x = -3/2, we have a minimum point.

Step 4: Substitute the found value of x into the original function to find the optimum value.
Substituting x = -3/2 into the original function f(x):

f(-3/2) = 2(-3/2)^2 + 6(-3/2) - 10
f(-3/2) = 2(9/4) - 18/2 - 10
f(-3/2) = 9/2 - 9 - 10
f(-3/2) = 9/2 - 18/2 - 20/2
f(-3/2) = -29/2

Therefore, the optimum value of the function f(x) = 2x^2 + 6x - 10 is -29/2 at x = -3/2, and it attains a minimum at that point.