Correct Answer - Option 2 : f(x) is minimum at x = - 2
Concept:
Let f be a function defined on an interval I.
First Derivative Test:
Calculate f’(x)
Solve f’(x) = 0 and find the roots. Suppose x = c is the root of f’(x) = 0.
Determine the sign of f’(x) for values of x slightly less than c and for values of x slightly greater than c.
If f’(x) changes sign from positive to negative as x increases through c then x= c is a point of maxima.
If f’(x) changes sign from negative to positive as x increases through c then x = c is a point of minima.
If f’(x) does not change sign as x increases through c, we say that x = c is neither a point of maxima nor minima.
In such case x = c is called the point of inflexion.
Similarly, we can say about the roots of f’(x) = 0 whether they are point of maxima or minima.
Second Derivative Test:
Calculate f’(x)
Solve f’(x) = 0 and find the roots. Suppose x = c is the root of f’(x) = 0.
Calculate f’’(x) and put x = c to get the value of f’’(c).
If f’’(c) < 0 then x = c is a point of maxima.
If f’’(c) > 0 then x = c is a point of minima.
If f’’(c) = 0 then we need to use the first derivative test.
Calculation:
Given, \(f\left( x \right) = {\left( {x + 2} \right)^2}\)
f’(x) = 2 (x + 2)
Now f’(x) = 0
⇒ x = - 2
Let us calculate f''(x) at this point.
f” (x) = 2
At x = - 2, f” (-2) = 2 > 0
According to second derivative test if f’’(c) > 0 then x = c is a point of minima
Hence, f(x) is minimum at x = - 2