Question
The point of local minimum & maximum of the function f (x) = x3 – (2x – 1)3 , is
Answers
GPT 3.5
to identify points on the graph where the function either reaches its highest point (maximum) or its lowest point (minimum) within a specific interval.
To find the local minimum and maximum of the given function f(x) = x^3 – (2x – 1)^3, we need to first calculate the derivative of the function, set it to zero, and solve for x.
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 3x^2 – 3(2x – 1)^2 * 2
Setting f'(x) = 0:
3x^2 – 12x^2 + 12x – 6 = 0
Combining like terms and simplifying:
-9x^2 + 12x – 6 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-B ± √(B^2 - 4AC)) / 2A
Here, A = -9, B = 12, and C = -6.
Plugging in the values, we get:
x = (-12 ± √(12^2 - 4(-9)(-6))) / 2(-9)
Simplifying further:
x = (-12 ± √(144 - 216)) / -18
x = (-12 ± √(-72)) / -18
Since the square root of a negative number is not a real value, there are no real solutions for x. Hence, there are no local minimum or maximum points for the function f(x) = x^3 – (2x – 1)^3.
To find the local minimum and maximum of the given function f(x) = x^3 – (2x – 1)^3, we need to first calculate the derivative of the function, set it to zero, and solve for x.
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 3x^2 – 3(2x – 1)^2 * 2
Setting f'(x) = 0:
3x^2 – 12x^2 + 12x – 6 = 0
Combining like terms and simplifying:
-9x^2 + 12x – 6 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-B ± √(B^2 - 4AC)) / 2A
Here, A = -9, B = 12, and C = -6.
Plugging in the values, we get:
x = (-12 ± √(12^2 - 4(-9)(-6))) / 2(-9)
Simplifying further:
x = (-12 ± √(144 - 216)) / -18
x = (-12 ± √(-72)) / -18
Since the square root of a negative number is not a real value, there are no real solutions for x. Hence, there are no local minimum or maximum points for the function f(x) = x^3 – (2x – 1)^3.