- #1

- 48

- 0

**1. The problem statement: Find f(x) given that f′′(x) = 3/squareroot of x, f(4) = 20 and**

f′(4) = 7.

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)

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)