Can You Find the Function Given Its Derivative and Initial Conditions?

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The discussion revolves around finding functions from their derivatives and initial conditions. Participants successfully derive functions from given derivatives, such as f'(x) = x^{-\frac{1}{3}} - 1, leading to f(x) = (3/2)x^{2/3} - x + 6. They also explore a projectile motion problem, confirming that integrating the acceleration function is necessary to determine velocity and height at various points. Additionally, they discuss modeling revenue functions through integration of the rate of change of revenue. The importance of including units in final answers is emphasized for clarity.
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Hello all

If we have f'(x) = x^{-\frac{1}{3}} - 1 and f(8) = 4 find f(x). Ok so f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + C. Since f(8) = 4 then f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + 6

If we have f''(x) = 2x^{2}, f'(3) = 10, f(3) = 6 find f(x). So we have f'(x) = \frac{2}{3}x^{3} - 8. Now f(x) = \frac{1}{6}x^{4} - 8x + C So would f(x) = \frac{1}{6}x^{4} - 8x + 16.5?



Thanks :smile:
 
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looks good :smile:
 
courtrigrad said:
Hello all

If we have f'(x) = x^{-\frac{1}{3}} - 1 and f(8) = 4 find f(x). Ok so f(x) = \frac{x^{\frac{2}{3}}}{\frac{2}{3}} - x + C. Since f(8) = 4,
f(x) = \frac{x^{\frac{2}{3}}}{\frac{2}{3}} - x + 6


It's correct.

Daniel.
 
And the second one is correct as well.

Daniel.
 
thanks a lot

Just had a few more questions:

An object is projected upwards from the ground with an iniital velocity of 80 feet per second.

(a) How long does it take the object to reach its maximum height?
(b) What is the maximum height?
(c) When is the velocity of the object half of its inital velocity?
(d) What is the height of the object when its velocity is one-half the initial velocity?

Ok so I assume that the acceleration is -32 \frac{ft}{sec} So for (a) v = -32x + C where v is the velocity. For all of these parts do I just take the integral to get back the original function and solve for the required variables?

Also if the rate of change of revenue ie modeled by \frac{dR}{dt} = 0.675t^{\frac{3}{2}}, 0\leq t\leq 225 then when t = 0, R = 0 if I want to find the model for the revenue function would i simply take the integral of it? Then to see if its at a particular value I just solve for it right?

Thanks :smile:
 
Pay attention with the first problem:it's v=v(t),not of "x".

Then for the second problem,yes,u'd have to integrate the derivative.

Daniel.
 
Yea I think I got it

(a) 2.5
(b) 100
(c) 1.25
(d) 75
 
Your answers look good. However, you should always try to present them with their UNITS. For instance, item "a" should be "2.5 sec".
 
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