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Relative minimum and maximum are the points where the function changes its direction from increasing to decreasing or vice-versa.$ \blacktriangleright$ If there is an open interval containing 'c' on which f(c) is a maximum, then is called a relative (or local) maximum of 'f'.

$\color{red} {OR}$ f(x) has relative maximum at x = c, if

$ \blacktriangleright$ If there is an open interval containing 'c' on which f(c) is a minimum, then is called a relative (or local) minimum of 'f'.

$\color{red} {OR}$ f(x) has relative maximum at x = c, if

Relative maximum at x= a and x= c so the points are (a,f(a)) and (c,f(c)).

Relative minimum at x = b so the point is (b,f(b)).

To find the extrema of a continuous function 'f' on a closed interval [a,b], use these steps

1) Find the derivative of the given function.

2) Make the derivative equal to zero and find the value of x in the interval (a,b).

3) Evaluate at each points number in the open interval(a,b).

4) Evaluate 'f' at each endpoint of [a,b].

5)The least of these values is the minimum. The greatest is the maximum.

f '(x) = $3x^{2} + 8x $

$3x^{2} + 8x $ = 0

x( 3x + 8) = 0

So, x =0 and x = -8/3

When x= 0, f(x) = $x^{3} + 4x^{2}$

f(0) = 0

When x= -8/3,

f(-8/3) = $(\frac{-8}{3})^{3} + 4(\frac{-8}{3})^{2}$

= $\frac{-512}{27} + \frac{256}{9}$

f(-8/3) =9.481

When x= -5,

f(-5) = $(-5)^{3} + 4(-5)^{2}$

= -125 +100

f(-5) = -25

From the above we can conclude that relative minimum at x = 0 and coordinate form is (0,0) and relative maximum at x = -8/3 and coordinate form is (-8/3,9.481).

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