let f(x) = ax^3 + 6x^2 + bx +4. determine the constants a and b so that f has a relative minimum at x = -1 and a relative maximum at x = 2. show workings

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llltkl | College Teacher | (Level 3) Valedictorian

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Here is a more direct approach for the same problem:

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

Differentiating with respect to x,

`f'(x)=3ax^2+12x+b`

Differentiating again with respect to x,

`f’’(x)=6ax+12`

f has a relative minimum at x=-1 and a relative minimum at x=2,

For a function to have a relative extremum (maximum/minimum) at a particular point, the first derivative at this point must be equal to zero and the second derivative should be less than zero for relative maximum and greater than zero for relative minimum.

Put the first derivatives at x=-1 and also at x=2 equal to zero to get two equations as follows:

`f’(-1)=3a(-1)^2+12(-1)+b=0`

`rArr 3a+b-12=0`

`rArr 3a+b=12` --- (i)

and

`f’(2)=3a(2)^2+12(2)+b`

`rArr 12a+b+24=0`

`rArr 12a+b=-24` --- (ii)

Subtract eq. (i) from eq. (ii),

`9a=-24-12=-36`

`rArr a=-36/9=-4`

Put this value of a in eq. (i) to get,

`3(-4)+b=12`

`rArr -12+b=12`

`rArr b=24`

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Check the values of second derivatives:

At x=-1, `f’’(-1)=6*(-4)*(-1)+12`

=36,

>0

At x=2, `f’’(2)=6*(-4)*(2)+12`

=-36,

<0

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Therefore, for a=-4 and b=24, the function `f(x)=ax^3+6x^2+bx+4` has a relative minimum at x=-1 and a relative maximum at x=2.

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`f(x)=ax^3+6x^2+bx+4 ` (a and b are two constants)

`f’(x)=3*ax^2+2*6x+b = 3ax^2+12x+b`

`f’’(x)=2*3ax+12=6ax+12`

f has a relative minimum at x=-1

therefore, at x=-1,  f’(x)=0 and f’’(x)>0

Put f’(x)=0 such that

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

`rArr 3a+b-12=0`

`rArr 3a+b=12 --- (i)`

Also, Put f’’(x)>0 such that

`6a(-1)+12gt0`

`rArr -6agt-12`

`rArr alt2`

Put this value of a in eq. (i), to get,

b>6

Therefore, the values of constants a and b, for which f has a relative minimum at x=-1 are a<2, b>6

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f has a relative maximum at x=2

therefore at x=2,  f’(x)=0 and f’’(x)<0

Put f’(x)=0 such that

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

`rArr 12a+b+24=0`

`rArr 12a+b=-24 --- (ii)`

Also, Put f’’(x)<0 such that

`6a(2)+12lt0`

`rArr 12alt-12`

`rArr alt-1`

Put this value of a in eq. (ii), to get,

b>-12

Therefore, the values of constants a and b, for which f has a relative maximum at x=2 are a<-1, b>-12

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