# Given: f(x) = 4x^3+ 21x^2–24x + 5Find (a) the relative extrema (maximum/minimum), (b) the largest interval(s) where the function is increasing, and (c) the largest open interval(s) where the...

Given: f(x) = 4x^3+ 21x^2–24x + 5

Find (a) the relative extrema (maximum/minimum), (b) the largest interval(s) where the

function is increasing, and (c) the largest open interval(s) where the function is decreasing. Be sure to show the sign graph of

f „(x).

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### 1 Answer

a) You need to find the first derivative of the function and then you should search for its roots such that:

`f'(x) = (4x^3+ 21x^2–24x + 5)'`

`f'(x) = 12x^2 + 42x - 24`

You need to solve the equation `12x^2 + 42x - 24 = 0` such that:

`12x^2 + 42x - 24 = 0`

You need to divide by 6 such that:

`2x^2 + 7x - 4 = 0`

You should use quadratic formula such that:

`x_(1,2) = (-7+-sqrt(49 + 32))/4`

`x_(1,2) = (-7+-sqrt(81))/4`

`x_(1,2) = (-7+-9)/4`

`x_1 = 2/4 =gtx_1 = 1/2`

`x_2 = -16/4 =gt x_2 = -4`

You should know that the values of the function are negative over `(-4,1/2)` and they are positive over `(-oo,-4) and (1/2,oo).`

**Hence, the function reaches its relative maximum at `x = -4` and it reaches the relative minimum at `x = 1/2` .**

**b) Since the values of first derivative are positive over`(-oo,-4) U(1/2,oo), ` hence, the function increases over `(-oo,-4) U(1/2,oo).` **

**c) Since the values of first derivative are negative over `(-4,1/2)` , hence, the function decreases over `(-4,1/2).` **

**Sources:**