Question

Find the minimum or maximum value of f(x) = x2 + 6x +11 .

Answer 1

Answer:

Therefore the Minimum value of f(x) is 2.

Step-by-step explanation:

Given:

$f(x)=x^{2} + 6x+11$

To Find:

minimum or maximum value of f(x)

Solution:

To find minimum or maximum value of f(x)

Step 1 . Find f'(x) and f"(x)

$f(x)=x^{2} + 6x+11$

Applying Derivative on both the side we get

$f'(x)=\dfrac{d(x^{2})}{dx}+\dfrac{d(6x)}{dx}+\dfrac{d(11)}{dx}$

$f'(x)=2x+6+0$

Again Applying Derivative on both the side we get

$f''(x)=\dfrac{d(2x)}{dx}+\dfrac{d(6)}{dx}$

$f''(x)=2$

Step 2. For Maximum or Minimum f'(x) = 0 to find 'x'

$2x+6=0\\\\2x=-6\\\\x=\dfrac{-6}{2}=-3$

Step 3. IF f"(x) > 0 then f(x) is f(x) is Minimum at x

            IFf"(x) < 0 then f(x) is f(x) is Maximum at x

Step 4. We have

$f''(x)=2$

Which is grater than zero

then f(x) is Minimum at x= -3

Therefore the Minimum value of f(x) is 2.