Question

F'(x)=0f(x)=x^3-6x^2+12x+1

Answer 1

f'(x)=3x^2-12x+12

f'(x)=0

3x^2-12x+12=0

3(x-2)^2=0

x=2

(x^n)'=nx^(n-1)

Answer 2

To find the critical points of the function $f(x) = x^3 - 6x^2 + 12x + 1$ and subsequently determine if they are maximum, minimum, or saddle points, we need to follow these steps:

1. Find the first derivative of the function ($f'(x)$).
2. Set $f'(x) = 0$ and solve for $x$ to find the critical points.
3. Use the second derivative test to determine the nature of each critical point (maximum, minimum, or saddle point).

Step 1: Find $f'(x)$ To find the first derivative of $f(x)$, we differentiate it with respect to $x$:

$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x + 1)$

Differentiating each term:

$f'(x) = 3x^2 - 12x + 12$

Step 2: Find the critical points To find the critical points, we set $f'(x) = 0$ and solve for $x$:

$3x^2 - 12x + 12 = 0$

Step 3: Determine the nature of critical points Now, we'll use the second derivative test to analyze each critical point.

The second derivative of $f(x)$ ($f''(x)$) is the derivative of $f'(x)$:

$f''(x) = \frac{d}{dx}(3x^2 - 12x + 12)$

Differentiating each term:

$f''(x) = 6x - 12$

Now, let's evaluate $f''(x)$ at each critical point.

1. Critical point $x_1$:

To find the value of $f''(x_1)$, we substitute $x_1$ into $f''(x)$:

$f''(x_1) = 6 \cdot x_1 - 12$

2. Critical point $x_2$:

To find the value of $f''(x_2)$, we substitute $x_2$ into $f''(x)$:

$f''(x_2) = 6 \cdot x_2 - 12$

Now, based on the second derivative test:

• If $f''(x_1) > 0$, and $f''(x_2) > 0$, then both critical points are minimum points.
• If $f''(x_1) < 0$, and $f''(x_2) < 0$, then both critical points are maximum points.
• If $f''(x_1)$ and $f''(x_2)$ have opposite signs, then there is a saddle point at one of the critical points.

Without the specific values of $x_1$ and $x_2$, I cannot give you the exact nature of each critical point. You would need to find the solutions for $x_1$ and $x_2$ by solving the quadratic equation $3x^2 - 12x + 12 = 0$, and then evaluate $f''(x)$ at those points to determine the nature of each critical point.