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If \(f\left( x \right) = {\left( {x + 2} \right)^2}\), then f(x)
1. f(x) is maximum at x = - 2
2. f(x) is minimum at x = - 2
3. f(x) is maximum at x = 2
4. f(x) is minimum at x = 2
5. None of these

1 Answer

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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

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