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.
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
To find the minimum value of the function h(x) = x - (12(x + 1)^(1/3) - 12) on the interval [0, 26], we can use the concept of calculus.
Step 1: Find the critical points.
Critical points occur where the derivative of the function is either zero or does not exist. In this case, we only need to check where the derivative equals zero. So, we find the derivative of h(x) with respect to x.
h'(x) = 1 - 4(x + 1)^(-2/3)
To find the critical points, we set h'(x) equal to zero and solve for x:
1 - 4(x + 1)^(-2/3) = 0
Now, we solve the equation for x:
4(x + 1)^(-2/3) = 1
(x + 1)^(-2/3) = 1/4
(x + 1)^(2/3) = 4
x + 1 = 4^(3/2)
x + 1 = 8
x = 7
So, we have one critical point at x = 7 within the given interval [0, 26].
Step 2: Evaluate the function at the critical point and endpoints.
To find the minimum value of h(x), we need to compare the function values at the critical point and the endpoints of the interval [0, 26].
h(0) = 0 - (12(0 + 1)^(1/3) - 12) = -12
h(7) = 7 - (12(7 + 1)^(1/3) - 12) = -3
h(26) = 26 - (12(26 + 1)^(1/3) - 12) = 25
So, we have h(0) = -12, h(7) = -3, and h(26) = 25.
Step 3: Determine the minimum value.
Since h(7) = -3 is the smallest function value among the critical point and endpoints, the minimum value of h(x) on the interval [0, 26] is -3.
Therefore, the minimum value of the function h(x) = x - (12(x + 1)^(1/3) - 12) on the interval [0, 26] is -3.