- SUPPOSE THAT F OF X IS A CONTINUOUS FUNCTION
WITH F OF 1 = -4 AND F OF 5 = 3.
DETERMINE WHICH CHOICE BEST DESCRIBES
THE FOLLOWING STATEMENT.
F OF X EQUALS 0 FOR SUM X IN THE CLOSED INTERVAL
FROM 1 TO 5.
AND OUR CHOICES ARE ALWAYS FALSE,
SOMETIMES TRUE AND SOMETIMES FALSE,
ALWAYS TRUE.
NOTICE THE CLOSED INTERVAL FROM 1 TO 5 COME
FROM THE INPUTS OF THE FUNCTION AND NOTICE HOW THE OUTPUTS,
OR THE FUNCTION VALUES, ARE OPPOSITES SIGNS.
ONE'S NEGATIVE AND ONE'S POSITIVE.
IF F OF 1 = -4, OUR CONTINUOUS FUNCTION
MUST PASS THROUGH THE POINT (1,-4)
AND IF F OF 5 = 3, IT MUST ALSO PASS THROUGH THE POINT (5,3).
SO THE KEY HERE IS THAT OUR FUNCTION IS CONTINUOUS
AND IT MUST PASS THROUGH THESE TWO POINTS.
LET'S TAKE A LOOK AT THIS ON A GRAPH.
IF OUR FUNCTION IS CONTINUOUS, THAT MEANS WE MUST BE ABLE
TO SKETCH IT, PASSING THROUGH THESE TWO POINTS
WITHOUT LIFTING UP OUR PENCIL.
SO IT SHOULD MAKE SENSE IF WE SKETCH
OUR CONTINUOUS FUNCTION, PASSING THROUGH
THESE TWO POINTS, IT MUST CROSS THE X AXIS
AT LEAST ONCE AND THE X AXIS IS WHERE THE FUNCTION VALUES
WOULD BE 0.
SO OUR FUNCTION MIGHT LOOK SOMETHING LIKE THIS.
NOTICE HERE, F OF 4 = 0 AND 4 WOULD BE
IN THIS CLOSED INTERVAL, BUT THERE MAY ACTUALLY
BE MORE THEN ONE X VALUE IN THIS INTERVAL
WHERE THE FUNCTION VALUE WOULD BE 0.
WE CAN SKETCH A CONTINUOUS FUNCTION
OVER THIS INTERVAL WITH MORE THEN ONE X INTERCEPT.
FOR EXAMPLE, THE FUNCTION MIGHT ALSO LOOK SOMETHING LIKE THIS.
AND NOTICE IN THIS CASE, THERE ARE THREE X VALUES
IN THIS CLOSED INTERVAL WHERE THE FUNCTION VALUE
WOULD BE 0.
HERE AT X = 2, 3, AND 4.
SO REGARDLESS OF HOW MANY X VALUES ARE IN THIS INTERVAL,
ESPECIALLY IF F OF X = 0, THERE MUST BE AT LEAST ONE
BECAUSE THE FUNCTION IS CONTINUOUS.
SO THE BEST CHOICE WOULD BE ALWAYS TRUE.
LET'S TAKE A LOOK AT SIMILAR QUESTION.
HERE WE ARE TOLD THAT F OF X IS A FUNCTION WITH F OF -1 = -3
AND F OF 6 = 2.
NOTICE HERE WE'RE NOT TOLD THE FUNCTION IS CONTINUOUS.
DETERMINE WHICH CHOICE BEST DESCRIBES
THE FOLLOWING STATEMENT.
AGAIN, F OF X = 0 FOR SUM X IN THE CLOSED INTERVAL
FROM -1 TO 6.
ONCE AGAIN, NOTICE THAT -1 AND 6 ARE THE GIVEN INPUTS
AND THE OUTPUTS, OR THE FUNCTION VALUES,
HAVE OPPOSITE SIGNS.
IF F OF -1 = -3, THEN THE FUNCTION
MUST CONTAIN THE POINT (-1,-3)
AND IF F OF 6 = 2, THE FUNCTION MUST ALSO CONTAIN
THE POINT (6,2).
LET'S TAKE A LOOK AT THIS ON THE GRAPH AS WELL.
AGAIN, WE'RE NOT TOLD THE FUNCTION IS CONTINUOUS.
IT MAY OR MAY NOT BE CONTINUOUS.
WE KNOW THAT IF IT IS CONTINUOUS
AND IT PASSES THROUGH THESE TWO POINTS,
THERE MUST BE AT LEAST SOME X VALUE IN THIS INTERVAL
WHERE THE FUNCTION VALUE WOULD BE 0.
FOR EXAMPLE, HERE THE FUNCTION VALUE WOULD BE 0 AT X = 3.
BUT WHAT IF THE FUNCTION IS NOT CONTINUOUS?
COULD WE SKETCH A NON-CONTINUOUS FUNCTION
OVER THIS CLOSED INTERVAL THAT DOES NOT PASS THROUGH
THE X AXIS?
AND THE ANSWER IS YES, WE COULD.
IT MIGHT LOOK SOMETHING LIKE THIS.
IT COULD PASS THROUGH THIS POINT HERE,
HAVE A BREAK IN THE GRAPH AT 4, AND THEN BEGIN AT 4 AGAIN
ABOVE THE X AXIS. LET'S SAY HERE.
NOTICE HOW WE DO HAVE A FUNCTION THAT PASSES THROUGH
THE GIVEN POINTS BASED UPON THE GIVEN FUNCTION VALUES,
BUT THERE IS NOT AN X VALUE SUCH THAT F OF X = 0,
WHICH MEANS, IN THIS CASE, OUR BEST CHOICE WOULD BE
IT COULD BE SOMETIMES TRUE AND SOMETIMES FALSE
BASED UPON WHETHER THE FUNCTION IS CONTINUOUS OR NOT.
SO THIS IDEA LEADS US TO WHAT IS CALLED
THE INTERMEDIATE VALUE THEOREM.
THE INTERMEDIATE VALUE THEOREM STATES THAT IF OUR FUNCTION F
IS CONTINUOUS ON A CLOSED INTERVAL
FROM "A" TO B AND K IS ANY REAL NUMBER
BETWEEN F OF "A" AND F OF B, THEN THERE IS AT LEAST
ONE NUMBER C IN THIS CLOSED INTERVAL
FROM "A" TO B SUCH THAT F OF C = K.
AS AN ILLUSTRATION, LET'S TAKE A LOOK AT THIS FUNCTION HERE.
LET'S CONSIDER THIS FUNCTION ON THE CLOSED INTERVAL FROM -3 TO 3
OR FROM HERE TO HERE.
LOOKING AT THIS POINT ON THE FUNCTION,
NOTICE F OF -3 = -2,
AND CONSIDERING THIS POINT ON THE FUNCTION WHEN X = 3,
NOTICE HOW F OF 3 = 2.
SO NOW WE'RE GOING TO PICK A VALUE K
BETWEEN F OF "A" AND F OF "B" OR IN THIS CASE BETWEEN -2 AND +2.
NOW, IN OUR PREVIOUS EXAMPLES WE CONSIDERED
THE FUNCTION VALUE OF 0, BUT IT DOESN'T HAVE TO BE
THE FUNCTION VALUE OF 0.
LET'S SAY WE CONSIDERED THE FUNCTION VALUE OF 1.
WELL, NOTICE HOW Y = 1 OR F OF X = 1
WOULD BE ANYWHERE ALONG THIS HORIZONTAL LINE.
AND NOTICE ON THE CLOSED INTERVAL
FROM -3 TO 3, THERE ARE ACTUALLY THREE X VALUES
IN THIS CLOSED INTERVAL WHERE THE FUNCTION VALUE WOULD BE 1.
THE FUNCTION VALUE IS 1 HERE, HERE, AND HERE.
AND THEREFORE THE VALUES OF C, OR THE VALUES OF X
IN THE CLOSED INTERVAL WITH A FUNCTION VALUE OF 1,
WOULD BE HERE, HERE, AND HERE.
AND THESE X VALUES ARE GUARANTEED
BY THE INTERMEDIATE VALUE THEOREM AS LONG
AS THE FUNCTION IS CONTINUOUS ON THE CLOSED INTERVAL
FROM "A" TO B, OR IN OUR CASE, FROM -3 TO 3.
I HOPE YOU FOUND THIS HELPFUL.