Answer:
2
Step-by-step explanation:
f(x) = x² + 6x + 11
= x² + 2(x)(3) + 3² - 3² + 11
= (x+3)² + 2
The minimum of the function will be at x=-3, and the value of the minimum will be equal to 2.
The minimum of the function is the value of function, that will be minimum. It is found by taking the derivative of the function.
To know the minimum value of the function, let's take the first derivative of the function,
[tex]f(x)=x^2+6x+11[/tex]
[tex]f'(x)=2x+6[/tex]
Substitute the value of derivative with 0, therefore,
[tex]f'(x)=2x+6=0\\\\2x+6=0\\\\x=-3[/tex]
Now, to know that the value of x=-3 will be minimum or maximum, find the second derivative of the function,
[tex]f'(x)=2x+6\\\\f''(x)=2[/tex]
As we can see that the value of the second derivative of the function is positive, therefore, at x=-3, the function will be minimum.
Further, substitute the value of x as 3 to know the minimum of the function,
[tex]f(x)=x^2+6x+11\\\\f(-3)=(-3)^2+6(-3)+11\\\\f(-3)=9-18+11\\\\f(-3)=2[/tex]
Hence, the minimum of the function will be at x=-3, and the value of the minimum will be equal to 2.
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