Can the Values of a Function be Determined Without Integrating?

[NotMrX]

Suppse the following function was written:

<br /> f(x)=\int_{0}^{x} \frac{t-1}{t^4+1} dt<br />

Then we could assume there is a solution:
f(x) = F(x) - F(0)

Take the derivative:
f'(x) = F'(x) - F'(0) = F'(x)
<br /> f&#039;(x)=\frac{x-1}{x^4+1}<br />

Then we could determine if the function is increasing or decreasing over an interval. Without taking the antiderivative how could we determine what the following values are:
f(0)
f(1)
f(-1)

 

[radou]

Suppse the following function was written:

<br /> f(x)=\int_{0}^{x} \frac{t-1}{t^4+1} dt<br />

Then we could assume there is a solution:
f(x) = F(x) - F(0)

F(x0) represents the value of the primitive function F at a point x0, not a function. Do not mess up variables x with fixed points, which are conventionally called x0, a, b, c, etc.

 

[HallsofIvy]

Suppse the following function was written:

<br /> f(x)=\int_{0}^{x} \frac{t-1}{t^4+1} dt<br />

Then we could assume there is a solution:
f(x) = F(x) - F(0)

Take the derivative:
f'(x) = F'(x) - F'(0) = F'(x)
<br /> f&#039;(x)=\frac{x-1}{x^4+1}<br />

Then we could determine if the function is increasing or decreasing over an interval. Without taking the antiderivative how could we determine what the following values are:
f(0)
f(1)
f(-1)

f(0) is easy:
F(0)= \int_0^0 \frac{t-1}{t^4-1}dx= 0
There is no way to determine f(1) or f(-1) without actually doing the integral.

 

[NotMrX]

f(0) is easy:
F(0)= \int_0^0 \frac{t-1}{t^4-1}dx= 0
There is no way to determine f(1) or f(-1) without actually doing the integral.

Thank you for your help.