Question

Find the minimum value of the following function:

h(x)= x-(12(x+1)^(1/3) -12) on the interval
[0, 26]

Honestly, I have found by graphing that the minimum value is at the coordinate point (9,-3), but I need to do this in a way that uses a more "calculus-like" method if you catch my drift.

I don't know how I would find the minimum value without graphing.

Answer 1

If your expression mean:

x - 12 * ∛( x + 1 ) - 12

then:

x - 12 * ∛( x + 1 ) - 12 = x - 12 * ( x + 1 ) ^ ( 1 / 3 ) - 12

If h' (x) = 0 h (x) has a local extrema ( maximum or minimum )

In this case:

h' (x) = 1 - 12 * ( 1 / 3 ) ( x + 1 ) ^ ( 1 / 3 - 1 ) =

1 - 12 / 3 ( x + 1 ) ^ ( - 2 / 3 ) =

1 - 4 ( x + 1 ) ^ ( - 2 / 3 ) =

1 - 4 * 1 / [ ∛( x + 1 ) ^ 2 ] =

1 - 4 / [ ∛( x + 1 ) ^ 2 ]


h' (x) = 0

1 - 4 / [ ∛( x + 1 ) ^ 2 ] = 0

- 4 = ( - 1 ) * [ ∛( x + 1 ) ^ 2 ] = - 1

- 4 = - [ ∛( x + 1 ) ^ 2 ] | Multiply both sides by - 1

4 = [ ∛( x + 1 ) ^ 2 ]

[ ∛( x + 1 ) ^ 2 ] = 4

√ [ ∛( x + 1 ) ^ 2 ] = √4

∛( x + 1 ) = ± 2

x + 1 = ± 2 ^ 3

x + 1 = ± 8

Equation has two solutions:

1.

x + 1 = - 8

x = - 8 - 1

x = - 9

2.

x + 1 = 8

x = 8 - 1

x = 7


Now you must do second derivative test.

If h (x)" > 0 then h (x) has a local minimum.

If h (x)" < 0 then h (x) has a local maximum.


h (x)" = [ 1 - 4 ( x + 1 ) ^ ( - 2 / 3 ) ] ' =

0 - 4 * ( - 2 / 3 ) * ( x + 1 ) ^ [ ( - 2 / 3 ) - 1 ] =

( 8 / 3 ) ( x + 1 ) ^ [ ( - 2 / 3 ) - 3 / 3 ] =

( 8 / 3 ) ( x + 1 ) ^ ( - 5 / 3 ) =

8 / [ 3 ∛( x + 1 ) ^ 5 ]

Now:

x = - 9

h" = 8 / [ 3 ∛( - 9 + 1 ) ^ 5 ] =

8 / [ 3 ∛- 8 ^ 5 ] =

8 / [ 3 * - 2 ^ 5 ] =

8 / [ 3 * - 32 ] =

8 / - 96 =

- 8 / 12 * 8 =

- 1 / 12 < 0

For x = - 9 h (x) has a local maximum.


x = 7

h" = 8 / [ 3 ∛( 7 + 1 ) ^ 5 ] =

8 / [ 3 ∛8 ^ 5 ] =

8 / [ 3 * 2 ^ 5 ] =

8 / [ 3 * 32 ] =

8 / 96 =

8 / 12 * 8 =

1 / 12 > 0

For x = 7 h (x) has a local minimum.


Minimum value of h (x)

h (min) = h (7) = 7 - 12 * ∛( 7 + 1 ) - 12 =

7 - 12 * ∛ 8 - 12 =

7 - 12 * 2 - 12 =

7 - 24 - 12 = - 29


h (min) = h (7) = - 29