Given f(x) = \frac{x^3-2x+5}{x+4} and f’(3) = \frac{a}{b}, where a and b are coprime positive integers, what is the value of a+b?
To find the value of a+b, we first need to find the derivative of the function f(x) and then evaluate it at x = 3.
1. Let's find the derivative of f(x) using the quotient rule:
- The quotient rule states that if we have a function u(x) = g(x)/h(x), where g(x) and h(x) are both differentiable functions, then the derivative of u(x) is given by:
u'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2
2. Applying the quotient rule to f(x):
- g(x) = x^3 - 2x + 5
- h(x) = x + 4
Let's calculate the derivatives of g(x) and h(x) separately:
- g'(x) = 3x^2 - 2 (differentiating x^3 - 2x + 5)
- h'(x) = 1 (differentiating x + 4)
Now, substituting g'(x), g(x), h'(x), and h(x) into the quotient rule formula:
f'(x) = ((3x^2 - 2) * (x + 4) - (x^3 - 2x + 5) * 1)/((x + 4)^2)
3. Simplifying the above expression:
f'(x) = (3x^3 + 12x^2 - 2x - 8 - x^3 + 2x^2 - 5)/((x + 4)^2)
= (2x^3 + 14x^2 - 2x - 13)/((x + 4)^2)
4. Now, let's evaluate f'(x) at x = 3 to find f'(3):
f'(3) = (2(3)^3 + 14(3)^2 - 2(3) - 13)/((3 + 4)^2)
= (2(27) + 14(9) - 6 - 13)/(49)
= (54 + 126 - 6 - 13)/(49)
= 161/49
The value of a is 161, and the value of b is 49. Since a and b are coprime positive integers, the value of a+b = 161 + 49 = 210.