Find f(x) such that f'(x) = sqrt x -1/ sqrt x and f(0) = 13.
ALSO if you know how...
find the antiderivative
(12x^7 + 3x^5/x^4) dx
thank you so much!
ALSO if you know how...
find the antiderivative
(12x^7 + 3x^5/x^4) dx
thank you so much!
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Problem 1:
f'(x) = x^(1/2) + x^(-1/2)
Integrate f'(x)dx to get f(x) = 2/3*x^(3/2) + 2*x^(1/2) + C1
Using your initial condition, plugging in 0 for x and 13 for f(x), you get that C1 = 13
Therefore, f(x) = 2/3*x^(3/2) + 2*x^(1/2) + 13
Problem 2:
I'm going to call the expression f'(x)
f'(x) = 12x^7 + 3x NOTE x^5/x^4 = x
Integrate f'(x)dx to get f(x) = 12/8*x^8 + 3/2*x^2 + C1
Simplifying: f(x) = 3/2*x^8 + 3/2*x^2 + C1
Therefore, f(x) = 3/2*(x^8 + x^2 + C) where C = 2/3*C1
f'(x) = x^(1/2) + x^(-1/2)
Integrate f'(x)dx to get f(x) = 2/3*x^(3/2) + 2*x^(1/2) + C1
Using your initial condition, plugging in 0 for x and 13 for f(x), you get that C1 = 13
Therefore, f(x) = 2/3*x^(3/2) + 2*x^(1/2) + 13
Problem 2:
I'm going to call the expression f'(x)
f'(x) = 12x^7 + 3x NOTE x^5/x^4 = x
Integrate f'(x)dx to get f(x) = 12/8*x^8 + 3/2*x^2 + C1
Simplifying: f(x) = 3/2*x^8 + 3/2*x^2 + C1
Therefore, f(x) = 3/2*(x^8 + x^2 + C) where C = 2/3*C1