1. ## Intervals

need help with few questions:

1) Find all open intervals on which $f(x)=\frac{x^2}{x^2+4}$ is decreasing.
this is what i did:
a. fount the derivative: $f'(x)=\frac{8x}{(x^2+4)^2}$
b. set it equal to zero: $8x=0$
c. $-{\infty} decreasing
is that all you do? I'm concern for this problem b/c they say to "Find all open intervals".

2) Consider $f(x)=x^{1/2}$. Find all value(s), c, in the interval [0,1] such that the slope of the tangent line to the graph of f at c is parallel to the secant line through the point (0, f(0)) and (1, f(1)).
$\frac{1}{2x^{1/2}}$ i have no idea what to do next...

3) Explain why $f(x)=\frac{1}{x}$ has a minimum on the interval [1,2] but not on the interval [-1,1].
graph:

is it b/c the line is decreasing at interval [1,2]?

2. Originally Posted by BryanK
need help with few questions:

1) Find all open intervals on which $f(x)=\frac{x^2}{x^2+4}$ is decreasing.
this is what i did:
a. fount the derivative: $f'(x)=\frac{8x}{(x^2+4)^2}$
b. set it equal to zero: $8x=0$
c. $-{\infty} decreasing
is that all you do? I'm concern for this problem b/c they say to "Find all open intervals".
To find where it is increasing you need to look at the sign of the derivative. Since $f'(x) = 8x/(x^2+4)^2$ and the denominator is positive the sign only depends on the numerator $8x$. Now if $x>0$ then the numerator is positive and so $f'(x)>0$ which means $f(x)$ is increasing. Similarly, $f(x)<0$ for $x<0$. So it is increasing on $(0,\infty)$ and decreasing $(-\infty,0)$.

3. um.. u sure that derivative is right?

i may be horribly mistaken but shouldn’t be:

x{squared}(x{squared}+4){-1}
= -x(x{squared}+4){-2}
:confused: