# Let f (x)= ax^3 + 6x^2 +bx + 4. Determine the constants a and b so f has a relative minimum at x= -1 and a relative maximum x= 2.

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To find the relative minimum and maximum of the equation `ax^3 + 6x^2 + bx + 4` we take the first derivative,` `

`3ax^2 + 12x + b`

Then substitute the value of relative minimum `x = -1`

`3a (-1)^2 + 12(-1) + b`

`3a - 12 + b`

`3a + b = 12` `larr` **Equation 1**

And substitute the value of relative maximum `x = 2`

`3a (2)^2 + 12(2) + b`

`12a + 24 + b`

`12 a + b = -24` `larr` **Equation 2**

So, we have two equations

`3a + b = 12`

`12a + b = -24`

Using elimination method, to solve for `a`

`-3a - b = -12`

`12a + b = -24`

`9a = -36`

`a = -4`

Then using the value `a = - 4` ,we substitute it in Equation 1

`3(-4) + b = 12`

`-12 + b = 12`

`b = 24`

Thus, the value of `a` is `-4` and `b` is `24`

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f(x) = ax^3 + 6x^2 +bx + 4

For relative maxima/minima, put first derivative as 0 ( slope is 0 at maxima/minima)

`dy/dx` f(x) = 3ax^2 + 12x+ b = 0

Given maxima and minima at x = -1 and x = 2, put values of x in the first derivative to get two equations.

3a(-1)^2 + 12(-1) + b = 0

i.e. 3a -12 + b = 0

similarly,

3a(2)^2 + 12(2) + b = 0

i.e. 12a + 24 + b = 0

solve for a and b, subtracting first equation from second we get

12a+ 24+ b - 3a + 12 - b = 0

9a = -36

a = -4

putting value of a in the equation to get value of b

12*-4 + 24 + b = 0

i.e b = 48-24 = 24.

Thus, a = -4 and b = 24