Antiderivatives (Is my book incorrect?)

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  • #1
01010011
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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)

 

Answers and Replies

  • #2
Doc Al
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Is my working correct, or is my math pdf (in red font) correct?
Your work is correct; the pdf is not. (You can check by working backwards and by plugging in the constraints and seeing if they are met.)
 
  • #3
01010011
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Your work is correct; the pdf is not. (You can check by working backwards and by plugging in the constraints and seeing if they are met.)

Thank you Doc Al.

I thought that since the pdf is from Portland University, it would have been reliable. However, it was still very helpful because it explained how to do antiderivatives.
Here is the link:
http://faculty.up.edu/wootton/Calc1/Section4.9.pdf" [Broken]
 
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