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
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.
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