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- WE WANT TO WRITE THE EXPONENTIAL FUNCTION F OF T
EQUALS 100 TIMES E RAISED TO THE POWER OF 0.2T
IN THE FORM G OF T EQUALS A TIMES B
RAISED TO THE POWER OF T.
IN BOTH EXPONENTIAL FUNCTIONS A REPRESENTS THE INITIAL VALUE
OR INITIAL AMOUNT.
SO IF THIS IS 100 HERE, A IS 100 IN THIS EQUATION AS WELL.
BUT FOR AN EXPONENTIAL FUNCTION,
WITH BASE E THE COEFFICIENT OF THE EXPONENT
OR IN THIS CASE
0.2 GIVES US THE CONTINUOUS EXPONENTIAL GROWTH
OR DECAY RATE
AND BECAUSE THIS IS POSITIVE
WE KNOW THIS FUNCTION MODELS A 20% CONTINUOUS GROWTH RATE.
IN G OF T,
B MINUS 1 IS GOING TO GIVE US THE EXPONENTIAL GROWTH
OR DECAY RATE PER UNIT OF TIME
SO IF T IS YEARS WE CAN USE B TO FIND THE EXPONENTIAL GROWTH
OR DECAY RATE PER YEAR.
SO THIS IS A CONTINUOUS GROWTH RATE
AND B WILL HELP US FIND THE GROWTH RATE PER UNIT OF TIME.
TO DO THIS CONVERSION, WE'LL USE TWO FUNCTION VALUES
OR TWO POINTS FROM F OF T.
WE ALREADY KNOW THE INITIAL VALUE IN BOTH CASES
IS EQUAL TO 100.
TO EMPHASIZE THIS LET'S GO AHEAD AND USE F OF ZERO.
F OF ZERO IS EQUAL TO 100 TIMES E TO THE ZERO,
WHICH IS 1.
SO THIS WOULD BE 100.
IF F OF ZERO EQUALS 100,
THIS WOULD CORRESPOND TO THE POINT WHERE THE X COORDINATE
OR IN THIS CASE THE T COORDINATE IS ZERO
AND THE Y COORDINATE IS 100.
LET'S ALSO USE F OF 1.
SO WE WOULD HAVE 100 TIMES E RAISED TO THE POWER OF 0.2.
WE'LL GET A DECIMAL APPROXIMATION HERE.
100 TIMES 3 RAISED TO THE POWER OF 0.2.
THIS WILL BE APPROXIMATELY 122.1403.
WHICH MEANS THIS REPRESENTS THE POINT
WHEN THE T COORDINATE IS 1 AND THE Y COORDINATE IS 122.1403.
NOW I CAN FIND G OF T.
AGAIN, WE ALREADY KNOW THAT A IS AN EQUAL ONE--
WE SHOULD HAVE ALREADY RECOGNIZED
THAT A IS EQUAL TO 100.
SO WE HAVE G OF T EQUALS 100 TIMES B
RAISED TO THE POWER OF T.
SO NOW WE'LL USE THE SECOND FUNCTION VALUE
OR SECOND POINT TO DETERMINE OUR BASE B.
G OF 1 MUST ALSO EQUAL 122.1403.
SO WE'LL HAVE G OF 1 EQUALS 100 TIMES OUR UNKNOWN BASE
RAISED TO THE POWER OF T WHICH IS 1,
EQUALS THE FUNCTION VALUE 122.1403.
NOW, BECAUSE THIS IS ROUNDED
WE ARE GOING TO HAVE A SLIGHT MARGIN OF ERROR
BUT WE CAN SEE THIS WILL BE 100 B EQUALS 122.1403.
DIVIDE BOTH SIDES BY 100.
WE HAVE B IS EQUAL TO 1.2214.
AGAIN IF YOU 'LL ROUND TO FOUR DECIMAL PLACES.
SO THAT MEANS G OF T, THE FUNCTION WE'RE TRYING TO FIND,
IS EQUAL TO 100 TIMES OUR BASE OF 1.2214
RAISED TO THE POWER OF T.
SO THIS FUNCTION TELLS US THAT IT WE TAKE THE BASE
AND SUBTRACT 1
THE GROWTH RATE WOULD BE APPROXIMATELY 22.14 PERCENT
PER UNIT OF TIME,
SO IF T WAS IN YEARS,
THIS WOULD BE A 22.14% GROWTH RATE PER YEAR
WHICH IS THE SAME AS A CONTINUOUS GROWTH RATE OF 20%.
TO VERIFY THAT OUR WORK IS CORRECT,
LET'S GO AHEAD AND COMPARE THE FUNCTION VALUES OF F OF T
AND G OF T USING THE TABLE OF THE GRAPHING CALCULATOR.
SO WE'VE ALREADY TYPED IN THE ORIGINAL FUNCTION IN Y1.
OF COURSE, WE HAVE TO USE X INSTEAD OF T
AND I'VE TYPED IN THE FUNCTION THAT WE FOUND IN Y2.
AND NOW WE'LL COMPARE THE FUNCTION VALUES
USING THE TABLE FEATURE.
SO I'LL PRESS SECOND GRAPH.
AND, NOW, THERE MIGHT BE A SLIGHT DIFFERENCE IN Y1 AND Y2
BECAUSE WE DID HAVE TO ROUND OUR BASE HERE.
BUT IF WE SCROLL DOWN WE CAN SEE
THAT THE VALUES OF Y1 AND Y2 ARE EQUAL TO EACH OTHER
THOUGH AT SOME POINT THERE IS GOING TO BE A SLIGHT MARGIN
OF ERROR.
BUT THIS IS MORE THAN ENOUGH INFORMATION TO VERIFY
THAT OUR WORK IS CORRECT.
I HOPE YOU FOUND THIS HELPFUL.