Consider the volume of a sphere, `V=4/3 pi r^3`. Suppose that you will with ballon with air
at a constant rate, `100" cm"^3/"s"`. How does the radius increase?
Consider the image below:
`f(x+h)~~f(x)+f'(x)*h` if h is "small."
Consider `f(x)=x^2-2x+2`
Consider `f(x)=x^3-x`
Consider `f(x)=3-x^2`
Consider `f(x)=x^3`
Consider `f(x)=(x^2-4x+5)/(x-2)`
A function is increasing if `f(x_2)>f(x_1)` whenever `x_2>x_1`
This means that `f(x_2)-f(x_1)>0` whenever `x_2-x_1>0`, that is both differences are positive.
Now, the secant slope between those points `(f(x_2)-f(x_1))/(x_2-x_1)` is also positive.
`lim_(x_2 to x_1)(f(x_2)-f(x_1))/(x_2-x_1)=f'(x_1)` so the derivative/slope of the function
is positve.
Similarly, for the function is decreasing, then `f'(x)<0`.
Given continuity and the intermediate value theorem, if `g(a)>0` and `g(b)<0` then there is a `c in (a,b)` where `g(c)=0`
The first derivative of a function can potentially locate extreme vales for us.
Let f be continuous over the closed interval `[a,b]` and differentiable over the open interval `(a,b)`.
Then, there exists at least one point `c in (a,b)` such that
`f'(c)=(f(b)−f(a))/(b−a)`.
Some important consequences of the mean value theorem are:
CONSTANT DIFFERENCE THEOREM
If `f` and `g` are differentiable over an interval `I` and `f′(x)=g′(x)` for all `x∈I`, then `f(x)=g(x)+C` for some constant `C`.
INCREASING AND DECREASING FUNCTIONS
Let `f` be continuous over the closed interval `[a,b]` and differentiable over the open interval `(a,b)`.
Consider `f(x)=2x^3-12x^2-72x+2` on the closed interval `[-5,7]`
Desmos Graph
The First Derivative Test
Suppose that `f` is a continuous function over an interval `I` containing a critical point `c`. If `f` is differentiable over `I`, except possibly at point `c`, then `f(c)` satisfies one of the following descriptions:
The Second Derivative: Desmos Example of Changing Slopes on a Curve.