Ex: Sketch the Graph of a Derivative Function Given the Graph of a Function


– WE WANT TO GRAPH THE FUNCTION THAT WOULD GIVE THE SLOPE
WITH THE TANGENT LINES TO THE GIVEN FUNCTION
GRAPHED HERE IN RED. WELL, THE FUNCTION THAT GIVES
THE SLOPE WITH THE TANGENT LINES WOULD BE THE DERIVATIVE
FUNCTION. SO WE’RE GIVEN F OF X.
WE WANT TO GRAPH F PRIME OF X. SO LOOKING AT THE GRAPH
OF F OF X, LET’S FIRST CONSIDER THE GRAPH
ON THE INTERVAL FROM NEGATIVE INFINITY TO +2 WHERE WE HAVE THIS SHARP POINT
IN THE GRAPH. SO, AGAIN, WE’LL CONSIDER
THE FUNCTION FIRST ON THIS INTERVAL HERE. WELL, THE INTERVAL
IS GOING TO BE OPEN ON 2, MEANING IT WON’T INCLUDE 2 BECAUSE NOTICE HOW
BECAUSE OF THIS SHARP POINT WE CAN’T SKETCH A TANGENT LINE, AND, THEREFORE, THE DERIVATIVE
FUNCTION WOULD BE UNDEFINED. SO, AGAIN, WE’RE CONSIDERING
THIS ON THE INTERVAL FROM NEGATIVE INFINITY TO +2,
OPEN ON +2. BECAUSE THE FUNCTION
IS LINEAR ON THIS INTERVAL, NOTICE IF WE SELECT ANY POINT ON
THE FUNCTION IN THIS INTERVAL, LET’S SAY THIS POINT HERE, THE TANGENT LINE WOULD CONTAIN
THIS LINEAR SEGMENT OF THE FUNCTION. FOR EXAMPLE, HERE’S THE TANGENT
LINE TO THE FUNCTION AT THIS POINT, AND NOTICE HOW IT CONTAINS
THIS FUNCTION ON THIS INTERVAL. WHICH MEANS, WE CAN USE
THE SLOPE OF THE FUNCTION ON THIS INTERVAL TO DETERMINE THE SLOPE
WITH A TANGENT LINE. SO IF WE SELECT TWO POINTS
ON THIS INTERVAL, LET’S SAY THIS POINT HERE
AND THIS POINT HERE, WE CAN USE THESE TWO POINTS TO DETERMINE THE SLOPE
OF THE FUNCTION, WHICH SHOULD ALSO BE THE SLOPE
WITH A TANGENT LINE. NOTICE TO GO FROM THIS FIRST
POINT TO THE SECOND POINT WE HAVE TO GO DOWN ONE UNIT, WHICH MEANS THE CHANGE OF Y
IS -1. AND THEN RIGHT ONE UNIT, WHICH
MEANS THE CHANGE OF X IS +1. WHICH MEANS, ON THIS INTERVAL THE SLOPES WITH THE TANGENT
LINES WOULD BE THE CHANGE IN Y
DIVIDED BY THE CHANGE OF X OR -1 DIVIDED BY 1 OR -1. SO NOW TO GRAPH
THE DERIVATIVE FUNCTION, WHICH, AGAIN, WE’LL GIVE THE
SLOPE WITH THE TANGENT LINES, WE’D HAVE THE CONSTANT FUNCTION
Y=-1 ON THE INTERVAL FROM NEGATIVE
INFINITY TO 2, OPEN ON 2. SO THE GRAPH WOULD LOOK
SOMETHING LIKE THIS WHERE AT 2 WE HAVE AN OPEN POINT BECAUSE THE DERIVATIVE
WOULD BE UNDEFINED. NOW LETS CONSIDER F OF X ON THE
INTERVAL FROM 2 TO INFINITY, AGAIN, OPEN ON 2. SO CONSIDERING THE FUNCTION ON
THE INTERVAL FROM 2 TO INFINITY, AND, AGAIN, NOTICE HOW IT IS
LINEAR ON THIS INTERVAL. SO IF WE SELECT ANY POINT ON THE
FUNCTION IN THIS INTERVAL, LET’S SAY THIS POINT HERE, ONCE AGAIN, THE TANGENT LINES
ARE GOING CONTAIN THE FUNCTION ON THIS INTERVAL. SO, ONCE AGAIN, WE CAN USE
THE SLOPE OF THE FUNCTION ON THIS INTERVAL TO DETERMINE THE SLOPES
WITH THE TANGENT LINES. SO LET’S GO AHEAD AND SELECT
TWO POINTS ON THIS INTERVAL. LET’S SAY THIS POINT HERE
AND THIS POINT HERE. NOTICE TO GO FROM
THE POINT ON THE LEFT TO THE POINT ON THE RIGHT WE HAVE TO GO UP TWO UNITS. SO THE CHANGE OF Y IS +2. AND THEN RIGHT ONE UNIT. SO THE CHANGE OF X IS +1. WHICH MEANS, THE SLOPES OF THE
TANGENT LINES ON THIS INTERVAL WOULD BE M=THE CHANGE OF Y
DIVIDED BY THE CHANGE OF X, OR 2 DIVIDED BY 1,
WHICH EQUALS 2. SO ON THE OPEN INTERVAL
FROM 2 TO INFINITY OUR DERIVATIVE FUNCTION WOULD BE
THE CONSTANT FUNCTION Y=2. WHICH MEANS, ON THIS INTERVAL THE DERIVATIVE FUNCTION
WOULD LOOK LIKE THIS. AN OPEN POINT AT X=2, AND THEN TO THE RIGHT OF 2
THE CONSTANT FUNCTION Y=2. SO, AGAIN,
FOR THIS PIECE OF THE FUNCTION THE SLOPES OF THE TANGENT LINES
WOULD BE -1. AND, AGAIN, WE CANNOT SKETCH
A TANGENT LINE AT X=2, SO OUR DERIVATIVE FUNCTION
IS UNDEFINED AT X=2. AND THEN TO THE RIGHT OF +2 ON THE OPEN INTERVAL FROM 2
TO INFINITY, FOR THIS PIECE HERE THE SLOPE OF THE TANGENT LINES
WOULD BE +2. I HOPE YOU FOUND THIS HELPFUL.  

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