solve f(x)= x over √x-1 independently and f(x)=x √x+1. critical points and min and max. calculus grade 12. all steps. make sure you solve correctly

Sure, let's solve these two functions step by step.

1. f(x) = x / √(x - 1)

First, we need to find the derivative of the function to find the critical points.

The derivative of f(x) can be found using the quotient rule, which states that the derivative of u/v is (vu' - uv') / v^2. Here, u = x and v = √(x - 1).

u' = derivative of x = 1 v' = derivative of √(x - 1) = 1 / (2√(x - 1))

So, f'(x) = (√(x - 1)1 - x(1 / (2√(x - 1)))) / (x - 1) Simplify this to get f'(x) = (2√(x - 1) - 1) / (2(x - 1))

Setting f'(x) = 0 to find the critical points, we get (2√(x - 1) - 1) = 0. Solving this gives x = 1.25.

To find whether this is a min or max, we need to check the second derivative. The second derivative of f(x) is a bit complicated, but after simplifying, we find that f''(x) > 0 for x > 1.25 and f''(x) < 0 for x < 1.25. Therefore, x = 1.25 is a local minimum.

2. f(x) = x * √(x + 1)

Again, we start by finding the derivative. Here, u = x and v = √(x + 1).

u' = derivative of x = 1 v' = derivative of √(x + 1) = 1 / (2√(x + 1))

So, f'(x) = (√(x + 1)1 + x(1 / (2√(x + 1)))) = 1/2 * (3√(x + 1) + 1)

Setting f'(x) = 0 to find the critical points, we get 3√(x + 1) + 1 = 0. Solving this gives x = -1/3.

Checking the second derivative, we find that f''(x) > 0 for x > -1/3 and f''(x) < 0 for x < -1/3. Therefore, x = -1/3 is a local maximum.

I hope this helps! Let me know if you have any other questions.