a. a) Graph the function. b. b) Draw tangent lines to the graph at point whose x -coordinates are –2, 0, and 1. c. c) Find f ' ( x ) by determining lim x → 0 f ( x + h ) − f ( x ) h . d. d) Find f ' ( − 2 ) , f ' ( 0 , ) and f ' ( 1 ) . These slopes should match those of the lines you drew in part ( b ). f ( x ) = 1 2 x − 3
a. a) Graph the function. b. b) Draw tangent lines to the graph at point whose x -coordinates are –2, 0, and 1. c. c) Find f ' ( x ) by determining lim x → 0 f ( x + h ) − f ( x ) h . d. d) Find f ' ( − 2 ) , f ' ( 0 , ) and f ' ( 1 ) . These slopes should match those of the lines you drew in part ( b ). f ( x ) = 1 2 x − 3
Solution Summary: The author illustrates the graph of the function f(x)=12x-3.
5. Find the arc length of the curve y = 3x³/2 from x = 0 to x = 4.
-6 -5
*
10
8
6
4
2
-2 -1
-2
1 2 3 4 5 6
-6
-8
-10-
The function graphed above is:
Concave up on the interval(s)
Concave down on the interval(s)
There is an inflection point at:
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Differential Equation | MIT 18.01SC Single Variable Calculus, Fall 2010; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=HaOHUfymsuk;License: Standard YouTube License, CC-BY