Antiderivatives (Is my book incorrect?)

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)

 
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01010011 said:
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.)
 
Doc Al said:
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"
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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