Question

1. Find a derivative of this function using chain rule f(x) = sqrt(1-x^2)

2. Find the two values of x for which the function f(x) = 4x^3 + 3x^2 - 6x + 1 has critical points. (local max and min)

3. Use second derivative test to find local min and max of the function f(x) = 1 + 3x^2 - 2x^3.

Answer 1

1. To find the derivative of f(x) = sqrt(1-x^2), we can use the chain rule:

f'(x) = -x / sqrt(1-x^2)

2. To find the critical points of f(x) = 4x^3 + 3x^2 - 6x + 1, we need to find the values of x where f'(x) = 0 or f'(x) is undefined. First, we find the derivative:

f'(x) = 12x^2 + 6x - 6

Setting f'(x) = 0, we get:

12x^2 + 6x - 6 = 0

Simplifying, we get:

2x^2 + x - 1 = 0

Using the quadratic formula, we get:

x = (-1 ± sqrt(1 + 8)) / 4

x = -1 or x = 1/2

So, the critical points are x = -1 and x = 1/2.

3. To use the second derivative test to find the local minima and maxima of f(x) = 1 + 3x^2 - 2x^3, we need to find the critical points and the second derivative:

f'(x) = 6x^2 - 6x

Setting f'(x) = 0, we get:

6x^2 - 6x = 0

Simplifying, we get:

6x(x - 1) = 0

So, the critical points are x = 0 and x = 1.

f''(x) = 12x - 6

At x = 0, f''(0) = -6, so f(x) has a local maximum at x = 0.

At x = 1, f''(1) = 6, so f(x) has a local minimum at x = 1.