The number f (a) is called the maximum value of function $f$ in I and the point a is called a point of maximum value of f in I.

(b) A function f is said to have a minimum value in I, if there exists a point a in I such that $f (a) \leq f (x) \; \forall x \in I$.

The number f (a), in this case, is called the minimum value of f in I and the point a, in this case, is called a point of minimum value of f in I.

Find the maxima and minima of Below function

$y=x^2$ for $x \in R$

Here is the graph of the function

we have f(x) = 0 if x = 0. Also $f (x) \geq 0$, for all $x \in R$.

Therefore, the minimum value of f is 0 and the point of minimum value of f is x = 0. Further, it may be observed from the graph of the function that f has no maximum value and hence no point of maximum value of f in R

The number f (a), in this case, is called an extreme value of f in I and the point a is called an extreme point.

- In last section we talked about the maximum and minimun values of the function,But there might be other points in the domain of the Function where it does not becomes maximum and minimum but it becomes maximum and minumum with respect to the neigbhouring point.
- These point are call Local Minima and Local Maximum
- Let f be a real valued function and let c be an interior point in the domain of f. Then

(a) c is called a point of local maxima if there is an h > 0 such that $f (c) > f (x), \; \forall x \in (c - h, c + h), x \ne c $

The value f (c) is called the local maximum value of f.

(b) c is called a point of local minima if there is an h > 0 such that $f (c) < f (x), \; \forall x \in (c - h, c + h), x \ne c$

The value f (c) is called the local minimum value of f .

- A function might have many local maxima and minima

(i) If $f'(x)$ changes sign from positive to negative as x increases through c, i.e., if $f'(x) > 0$ at every point sufficiently close to and to the left of c, and $f'(x) < 0$ at every point sufficiently close to and to the right of c, then c is a point of

(ii) If $f'(x)$ changes sign from negative to positive as x increases through c, i.e., if $f'(x) < 0$ at every point sufficiently close to and to the left of c, and $f'(x) > 0$ at every point sufficiently close to and to the right of c, then c is a point of

(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection

if \( f'(x) \) changes from positive to negative as x increase from left to right of that point,then that point is Local Maxima

if \( f'(x) \) changes from negative to positive as x increase from left to right to that point , then that point is Local Minima

(i) x = c is a point of local maxima if $f'(c) = 0$ and $f''(c) < 0$ The value f (c) is local maximum value of function $f$ .

(ii) x = c is a point of local minima if $f'(c) = 0$ and $f''(c) > 0$ In this case, f (c) is local minimum value of function $f$ .

(iii) The test fails if $f'(c) = 0$ and $f''(c) = 0$. In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion

- If \( f''(x) > 0 \) at a point where \( f'(x) = 0 \), \( f(x) \) has a local minimum at that point.

(i) then we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion

- Local Maxima: \( f'(x) \) changes from positive to negative.

- Local Minima: \( f'(x) \) changes from negative to positive.

(ii) Find the third derivative and if \( f'''(a) \ne 0 \), then this point is point of inflexion

If \( f'''(a) = 0 \),then find fourth derivative,if \( f''''(a) < 0 \) then point a is local maximum and if \( f''''(a) > 0 \) then point a is local minimum

- This the maximum value of the function in the closed interval
- This is also called global maximum or greatest value

- This the minimum value of the function in the closed interval
- This is also called global minimum or least value.

Here f(a) : Absolute Minimum in the closed Interval

f(b) : Local Maxima

f(c): Local Minima

f(d): Absolute Maximum

Find the maxima and minima of the function \(f(x) = x^3 - 3x^2 - 9x + 5\).

Step 1: Find critical points \[f'(x) = 3x^2 - 6x - 9\]

Set \(f'(x) = 0\)\

\[3x^2 - 6x - 9 = 0\]

\[x^2 - 2x - 3 = 0\]

\[x = 3, x = -1\]

Step 2: Use the second derivative test.

\[f''(x) = 6x - 6\]

For \(x = 3\), \(f''(3) = 12 > 0\). This means \(x = 3\) is a point of minima.

For \(x = -1\), \(f''(-1) = -12 < 0\). This indicates \(x = -1\) is a point of maxima.

Determine the maxima and minima of the function \(f(x) = x^2 e^{-x}\) for \(x \geq 0\).

Step 1: Find critical points

\[f'(x) = (2x - x^2) e^{-x}\]

Set \(f'(x) = 0\).

\[2x - x^2 = 0\]

\[x = 0, x = 2\]

Step 2: Second derivative test.

\[f''(x) = (x^2 - 4x + 2) e^{-x}\] For \(x = 0\), \(f''(0) = 2 > 0\). Minima.

For \(x = 2\), \(f''(2) = -4e^{-2} < 0\). Maxima.

Therefore

The function has a maximum at \(x = 2\) and a minimum at \(x = 0\).

**Notes**-
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