Expert: Scotto - 10/25/2009

Question
My Problems (set 1):
1. In Exercise 1, a function y=f(x) and values x0 and x1 are given.
(a) Find the average rate of change of y with respect to x over the interval [x0, x1].
(b) Find the instantaneous rate of change of y with respect to x at the given value of x0.
(c) Find the instantaneous rate of change of y with respect to x at a general point x0.
This is what is given --- y=(1/2x²); x0=3, x1=4
7. In Exercise 7, a function f and a value of x0 are given.
(a) Find the slope of the tangent to the graph of f at a general point x0.
(b) Use the result in part (a) to find the slope of the tangent at the given value of x0.
This is what is given --- f(x)=radical(x); x0=1.
12. In Exercise 12, use the limit process to find the slope of the graph of the function at the specified point.
This is what is given --- h(x)=(2x+5), (-1, 3)
22. In Exercise 22, find the formula for the slope of the graph. Use the formula to find the slope at the two points.
This is what is given --- g(x) =(x³) (a) (1, 1) (b) (-2, -8)

Answer
1. In Exercise 1, a function y=f(x) and values x0 and x1 are given.
(a) Find the average rate of change of y with respect to x over the interval [x0, x1].
The slope is the rate of change and is given as (y1-y0)/(x1-x0) or (f(x1)-f(x0))/(x1-x0)

(b) Find the instantaneous rate of change of y with respect to x at the given value of x0.
That would be f'(x) for the given x.

(c) Find the instantaneous rate of change of y with respect to x at a general point x0.
That would be f'(x0).

This is what is given --- y=(1/2x²); x0=3, x1=4
7. In Exercise 7, a function f and a value of x0 are given.
(a) Find the slope of the tangent to the graph of f at a general point x0.
Tangent slope is the derivative where it touches the graphl.
For this equation, f(x) = x²/2, so f'(x) = 2x/2 = x.
So at x0, the slope is x0.

(b) Use the result in part (a) to find the slope of the tangent at the given value of x0.
This is what is given --- f(x)=radical(x); x0=1.
Is that f(x)=√x?  Then that is f(x) = x^0.5, so f'(x) = 0.5/x^0.5.
With x0=1, 0.5/1^0.5 = 0.5/1 = 0.5.

12. In Exercise 12, use the limit process to find the slope of the graph of the function at the specified point.  This is what is given --- h(x)=(2x+5), (-1, 3)
Since h(x) is a line, the slope is always the same.
The slope of a line is given by the factor in front of x.
Here, the slope is 2.

22. In Exercise 22, find the formula for the slope of the graph. Use the formula to find the slope at the two points.  This is what is given --- g(x) =(x³) (a) (1, 1) (b) (-2, -8)
The slope of a line is g'(x).  If g(x) = x³, then g'(x) = 3x².
At (1,1), the slope is 3•1² = 3•1 = 3.
At (-2,-8), the slope is 3•(-2)² = 3 •4 = 12.