Determine whether this function has a mzximum or minimum value and then find that value:
f(x)=2x^2 + 3x -9
The derivative, f'(x) = 4x +3, is zero when x = -3/4 . That means there is a relative extreme value (either maximum or minimum) there.
The second derivative is f''(x)= 4 everywhere. That means its extreme value at x = -3/4 is a minimum.
To find the maximum or minimum value of the function f(x) = 2x^2 + 3x - 9, we need to find the x-value where it occurs.
Here's how you can do it step by step:
1. Take the derivative of the function f(x) with respect to x. This will give you f'(x), which represents the slope of the function at any given point.
f'(x) = 4x + 3
2. Set f'(x) equal to zero and solve the equation for x. This will give you the x-value(s) where the slope is zero.
4x + 3 = 0
4x = -3
x = -3/4
So, the x-value where the function may have a maximum or minimum value is x = -3/4.
3. To determine whether it is a maximum or minimum value, you need to look at the concavity of the function, which is determined by the second derivative.
Take the derivative of f'(x), which is the second derivative of f(x), denoted as f''(x).
f''(x) = 4
Since the second derivative f''(x) is positive everywhere (4 is always positive), it indicates that the function is concave up, and the x-value -3/4 represents a relative minimum value.
4. Finally, substitute the x-value (-3/4) back into the original function f(x) to find the y-value (maximum or minimum value).
f(-3/4) = 2(-3/4)^2 + 3(-3/4) - 9
f(-3/4) = 2(9/16) - (9/4) - 9
f(-3/4) = 9/8 - 9/4 - 9
f(-3/4) = -69/8
Therefore, the function f(x) = 2x^2 + 3x - 9 has a relative minimum value of -69/8 when x = -3/4.