Find the Derivative of a Quadratic Function with f(0)=-4 and Evaluate f'(0)

[ProBasket]

Let $$f(x)$$ be a quadratic function such that $$f(0) = -4$$ and

$$\frac{f(x)}{x^2(x-5)^8}dx$$
is a ration function.

Determine the value of $$f'(0)$$.

$$f'(0)=______$$

i don't really have a clue on how to do this. I can only think of integrating the function then find the derivative of it and plug in 0 for the x's, which doesn't seem to be correct.

 

[xanthym]

Let $$f(x)$$ be a quadratic function such that $$f(0) = -4$$ and

$$\frac{f(x)}{x^2(x-5)^8}dx$$
is a ration function.

Determine the value of $$f'(0)$$.

$$f'(0)=______$$

i don't really have a clue on how to do this. I can only think of integrating the function then find the derivative of it and plug in 0 for the x's, which doesn't seem to be correct.

Because the integrand's numerator is a polynomial of lesser order than the denominator, we can use partial fraction decomposition to obtain:
f(x)/{(x^2)(x-5)^8} =
= a_1/x + a_2/x^2 + b_1/(x - 5) + b_2/(x - 5)^2 + b_3/(x - 5)^3 + ... + b_8/(x - 5)^8

Since the integral is given to be purely rational, {a_1=0} and {b_1=0} since otherwise LOG terms would result. Thus, adding remaining terms on the right with common denominator of {(x^2)(x-5)^8}, we can equate f(x) with the numerator:
f(x) = {a_2*(x-5)^8} + {b_2*(x^2)*(x-5)^6} + {b_3*(x^2)*(x-5)^5} + ... + {b_8*(x^2)}

Because it's given that f(0)=(-4), we have:
f(0) = (-4) = {a_2*((0) - 5)^8} + 0 + 0 + ... + 0
(-4) = a_2*(5^8)
a_2 = (-4)/(5^8)

Furthermore, f'(x) will have the form:
f'(x) = (8)*(a_2)*(x-5)^7 + {terms involving either (x) or (x^2)}
so that substituting x=(0) and a_2={(-4)/(5^8)} from above:
f'(0) = (8)*{(-4)/(5^8)}*{(0) - 5)^7} + 0 + 0 + ... + 0
f'(0) = (8)*(4)*(5^7)/(5^8)
f'(0) = (32/5)


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[wais]

To find the derivative of a quadratic function, we can use the power rule. Let's first rewrite the given function as:

f(x) = -4x^2(x-5)^-8

Using the power rule, we can find the derivative as:

f'(x) = -8x(x-5)^-9 + 2(-4x)(x-5)^-8

Now, to find the value of f'(0), we can plug in x=0 into the derivative function:

f'(0) = -8(0)(0-5)^-9 + 2(-4)(0)(0-5)^-8

Simplifying, we get:

f'(0) = 0 + 0 = 0

Therefore, the value of f'(0) is 0. This makes sense since the derivative of a quadratic function at its vertex is always 0.