- #1
01010011
- 48
- 0
Is my working correct, or is my math pdf (in red font) correct?
1. The problem statement: Find f(x) given that f′′(x) = 3/squareroot of x, f(4) = 20 and
f′(4) = 7.
2. Most General Antiderivative: F(X) + C
3. The Attempt at a Solution
First, simplify 3/squareroot of x:
= 3x^(-1/2)
Next, most general antiderivative = F(x) + C
= [3x^(-1/2 + 1)] / [-1/2 + 1] + C
= [3x^1/2] / [1/2] + C
= 2 * 3x^1/2 + C
= 6x^1/2 + C
Now, f′(4) = 6x^1/2 + C = 7
= 6(4)^1/2 + C = 7 (The pdf has: f′(4) = 6 · 2 + C = 7 so C = −1. )
= 6(2) + C = 7
= 12 + C = 7
C = 7 - 12
C = -5
= f′(x) = 6x^(1/2) − 5 (The pdf has:f′(x) = 6x^(1/2) − 1. )
= [6x^(3/2)] / (3/2) - 5x + C
= 2/3 * 6x^ 3/2 - 5x + C
= 4x^3/2 -5x + C
f(4) = 4(4)^3/2 - 5(4) + C = 20
4 * 8 - 20 + C = 20
C = 20 - 32 + 20
C = 8
Therefore, f(x) = 4x^3/2 - 5x + C (The pdf has:f(x) = 4x^(3/2)− x)
1. The problem statement: Find f(x) given that f′′(x) = 3/squareroot of x, f(4) = 20 and
f′(4) = 7.
2. Most General Antiderivative: F(X) + C
3. The Attempt at a Solution
First, simplify 3/squareroot of x:
= 3x^(-1/2)
Next, most general antiderivative = F(x) + C
= [3x^(-1/2 + 1)] / [-1/2 + 1] + C
= [3x^1/2] / [1/2] + C
= 2 * 3x^1/2 + C
= 6x^1/2 + C
Now, f′(4) = 6x^1/2 + C = 7
= 6(4)^1/2 + C = 7 (The pdf has: f′(4) = 6 · 2 + C = 7 so C = −1. )
= 6(2) + C = 7
= 12 + C = 7
C = 7 - 12
C = -5
= f′(x) = 6x^(1/2) − 5 (The pdf has:f′(x) = 6x^(1/2) − 1. )
= [6x^(3/2)] / (3/2) - 5x + C
= 2/3 * 6x^ 3/2 - 5x + C
= 4x^3/2 -5x + C
f(4) = 4(4)^3/2 - 5(4) + C = 20
4 * 8 - 20 + C = 20
C = 20 - 32 + 20
C = 8
Therefore, f(x) = 4x^3/2 - 5x + C (The pdf has:f(x) = 4x^(3/2)− x)