Consider the function f(x) = x + 2/x, defined over the domain 1 ≤ x ≤ 3. Draw its graph by following the steps: a) Find f′(x); b) Solve f′(x) = 0 c) Find where the function is increasing and decreasing d) Find f′′(x) e) Find where the function is concave up and concave down f) Find the coordinates of all points of interest
I already have part a and b but im stuck on c
The function is a line. It has positive slope. It is increasing over all intervals.
is there a way i can prove that mathematically?
Draw the graph
awesome thanks. i can do part d but what about part e
Also if the first derivative is positive, the function is increasing.
If the second derivative is positive, it is concave up
the whole function or only some of it?
On the interval where it is positive
f(x) is concave up on I iff f''(x) \>= 0$ on I.
and to find the coordinates in part f i would use the original function and only include the numbers that are within the domino right?
*domain
do i just make x+(2/x)=0 and find values for x?
Where is x + (2/x)=0 coming from?
the orig function
is that the right way to do it. i dont know how to find the intersect if its a straight line??
That's not the original function you posted.
dang typo. it was meant to be F(x)=x+(2/x)
i did part a and be right and the fist derivative is positive so it is increasing. the second derivative is positive also so it is concave up so what steps do i need to take for finding the coordinates of all points of intersect
What did you get for \(F'(x)\)?
1-(2/x^2)
ok...so on what interval is \(F\) increasing?
Mertsj said it was increasing over all intervals
that was your previous function
ohhhh so is it sill increasing?
you tell me
ok so if the derivative is f'(x)=1-(2/x2), then would it mean that it decreases because the x in the equation is negative? ?
Ive worked out hat for part b) when f'(x) = 0 , x=plus and minus the square root of 2. Is that right?
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