# Math Help - Finding f(x)

1. ## Finding f(x)

Let f be the function defined for all x > -5 and having the following properties. Find an expression for f(x).
i) f ''(x) = 1/ (x+5)^1/3 for all x in the domain of f
ii) the line tangent to the graph of f at (4,2) has an angle of inclination of 45 degress.

I know

f '(x) =(3/2) (x+5)^2/3 + C
f (x) =(9/10) (x+5)^5/3 + (about -5.49)x + C

Well solving for C in f'(x) I got about -5.49
Solving for C to get f(x) I got about 70.254

Is this correct? If it is, is there an easier way to express those numbers in exact terms?

2. This is what i'm getting.

$\int f''(x)dx = f'(x) = \int (x + 5)^{-1/3}dx = \frac {3}2{(x + 5)^{2/3}} + c$

f'(4) = 1.

$1 = \frac {3}2(4 + 5)^{2/3} + c$

$1 = \frac {3}2(9)^{2/3} + c$

$c = 1 - (2187/8)^{1/3}$

$f(x) = \int (\frac {3}2{(x + 5)^{2/3}} + 1 - (2187/8)^{1/3})dx$

3. Do you know how to simplify the calculation for the second C?

4. I personally don't find the big fractions to look any better. And they certainly don't make it "easier" to express =p

$\int (\frac {3}2{(x + 5)^{2/3}} + 1 - (2187/8)^{1/3})dx = \frac{9}{10}(x + 5)^{5/3} + x - (2187/8)^{1/3}x + c$

$2 = \frac{9}{10}(4 + 5)^{5/3} + 4 - (2187/8)^{1/3}4 + c$

$c = 6 - (43046721 / 1000)^{1/3} - (139968/8)^{1/3}$

To actually shorten that further would be more effort than it's worth. 0.o

5. Yeah I personally don't like them either but my teacher is a real stingy about getting exact answers. thanks!

$c = 6 - (43046721 / 1000)^{1/3} - (139968/8)^{1/3}$

$c = 6 - \frac {43046721^{1/3}}{1000^{1/3}} - \frac {139968^{1/3}}{8^{1/3}}$

$c = 6 - \frac {43046721^{1/3}}{10} - \frac {139968^{1/3}}{2}$

$c = 6 - \frac {243(3)^{1/3}}{10} - \frac {36(3)^{1/3}}{2}$

$c = 6 - \frac {486(3)^{1/3}}{20} - \frac {360(3)^{1/3}}{20}$

$c = 6 - \frac {63(3)^{1/3}}{10}$

Assuming my original arithmetic was at all correct.

7. And here I get another answer =p Guess i'm really bad at this.

$\frac{9}{10}(x + 5)^{5/3} + x - \frac{(2187)^{1/3}x}{2} + c$

$2 = \frac{9}{10}(9)^{5/3} + 4 - 2(2187)^{1/3} + c$

$c = 18(3)^{1/3} - \frac{243(3)^{1/3}}{10} - 2$

$c = \frac{-63(3)^{1/3}}{10} - 2$