Consider the function:𝑓(𝑥)=𝑥3−3𝑥22−36𝑥Find the relative maximum point on this function.

To find the relative maximum point of the function f(x) = x^3 - 3x^2 - 22x, we first need to find the derivative of the function.

The derivative of the function f(x) is f'(x) = 3x^2 - 6x - 22.

Next, we set the derivative equal to zero and solve for x to find the critical points:

3x^2 - 6x - 22 = 0

This is a quadratic equation, and we can solve it using the quadratic formula:

x = [6 ± sqrt((6)^2 - 43(-22))] / (2*3) x = [6 ± sqrt(36 + 264)] / 6 x = [6 ± sqrt(300)] / 6 x = [6 ± 10sqrt(3)] / 6 x = 1 ± sqrt(10/3)

So the critical points are x = 1 - sqrt(10/3) and x = 1 + sqrt(10/3).

To determine whether these points are relative maxima, minima, or neither, we can use the second derivative test. The second derivative of the function is f''(x) = 6x - 6.

We substitute the critical points into the second derivative:

f''(1 - sqrt(10/3)) = 6(1 - sqrt(10/3)) - 6 = 6 - 6sqrt(10/3) - 6 = -6sqrt(10/3) f''(1 + sqrt(10/3)) = 6(1 + sqrt(10/3)) - 6 = 6 + 6sqrt(10/3) - 6 = 6sqrt(10/3)

The second derivative is negative at x = 1 - sqrt(10/3) and positive at x = 1 + sqrt(10/3), so by the second derivative test, the function has a relative maximum at x = 1 - sqrt(10/3).

To find the y-coordinate of the relative maximum, we substitute x = 1 - sqrt(10/3) into the original function:

f(1 - sqrt(10/3)) = (1 - sqrt(10/3))^3 - 3(1 - sqrt(10/3))^2 - 22(1 - sqrt(10/3))

So the relative maximum point on the function is (1 - sqrt(10/3), f(1 - sqrt(10/3))).