Caclulus Help!?

Question

My book says Find all relative extrema. Use the second derivative test where applicable. Heres my first problem
f(x)=6x-x^2 i have no idea how to start

Answer 1

The relative extrema exist where the slope is zero.

The slope is zero where the first derivative is zero.

f'(x)=6-2x
f'(x)=0 if x=3

If f''(x) < 0 then f has a maximum at x.
If f''(x) > 0 then f has a minimum at x.
Note that if f''(x) = 0 the second derivative test says nothing about the point x.

f''(x)= -2
f''(3)= -2

So, f(x)=6x-x^2, has a maximum at x=3.

I'd graph the function to see that the answer makes sense.

Answer 2

A function will give you a position of y for each point x.
A first derivative will give you the slope of the line for each point x.
The second derivative will give you where the slope is not changing.

You can consider this as position, speed, and acceleration.

If you are trying to find where the local minima and maxima are, just set the first derivative to zero.

The first derivative of f(x) is 6-2x
Setting this to zero, you get:
6-2x=0
6 = 2x
3 = x
So, at x = 3, there is a minima or maxima.

The second derivative of f(x) is the derivative of 6-2x which is -2
Therefore, as x increases, the upward direction of y keep going down. This means that y reaches its maximum at 3 then plummets. It is a parabola shape.

Answer 3

Why are you taking Calculus!
I heard that Calculus is extremely difficult.
I'm only a freshman, so I don't have that class...yet.

Answer 4

Calculus is the devil. I feel your pain.

Answer 5

calculus? wow, seriously now.......