 – WE WANT TO FIND THE DOMAIN
OF THE SQUARE ROOT FUNCTION F OF X=THE SQUARE ROOT OF THE
QUANTITY -X SQUARED + 4X + 5. THE DOMAIN OF THE FUNCTION WILL BE THE SET OF ALL POSSIBLE
INPUTS OR ALL POSSIBLE X VALUES
IN THIS CASE. WHEN CONSIDERING THE SQUARE ROOT
FUNCTION, WE NEED TO RECOGNIZE
THAT THE RADICAND OR THIS EXPRESSION
UNDERNEATH THE SQUARE ROOT CANNOT BE NEGATIVE, OTHERWISE THE FUNCTION VALUE
WOULD NOT BE REAL. SO OUR RESTRICTION
IS THIS QUANTITY HERE MUST BE GREATER THAN
OR EQUAL TO ZERO. SO BY SOLVING THIS QUADRATIC
INEQUALITY, WE’LL BE DETERMINING THE DOMAIN
OF OUR FUNCTION. SO TO DETERMINE
WHEN -X SQUARED + 4X + 5 IS GREATER THAN OR=0, WE’LL FIRST DETERMINE
THE X VALUES WHERE THIS IS EQUAL TO ZERO. AND THEN WE’LL USE THOSE VALUES TO FORM INTERVALS ON THE NUMBER
LINE, AND THEN TEST TO SEE WHICH INTERVALS SATISFY THIS
ORIGINAL INEQUALITY. SO, AGAIN, THE FIRST STEP
IS TO DETERMINE WHEN -X SQUARED + 4X + 5=0. THIS IS FACTORABLE. LET’S GO AHEAD AND FACTOR
OUR NEGATIVE OR -1 SO WE HAVE AN LEADING
COEFFICIENT THAT’S POSITIVE. THIS WILL CHANGE THE SIGN
OF EACH TERM, SO WE’LL HAVE
X SQUARED – 4X – 5=0. SO WE’LL FACTOR
INTO TWO BINOMIAL FACTORS. NOW X AND X, THE FACTORS OF -5
ADDED TO -4 WOULD BE -5 AND 1. SO THIS IS=0
WHEN X=5 AND X=-1. LOOKING BACK AT THE ORIGINAL
INEQUALITY FOR A MOMENT, WE WANT TO KNOW
WHEN THIS EXPRESSION IS GREATER THAN OR EQUAL
TO ZERO. WELL, THESE THE TWO VALUES WHERE
IT’S EQUAL TO ZERO, SO THESE ARE TWO SOLUTIONS
OR TWO VALUES IN OUR DOMAIN. BUT NOW WE’LL ALSO USE
THESE VALUES TO FORM INTERVALS
IN THE NUMBER LINE TO SEE WHICH INTERVALS SATISFY
THIS INEQUALITY. SO BECAUSE THESE DO SATISFY
THE ORIGINAL INEQUALITY, WE’RE GOING TO MAKE CLOSE POINTS
AT -1 AND 5. IF THIS WAS ONLY GREATER THAN 0, WE WOULD ACTUALLY MAKE OPEN
POINTS ON THESE TWO VALUES. NOW WE’RE GOING TO PICK TEST
VALUES IN EACH INTERVAL TO DETERMINE WHICH INTERVALS
SATISFY THIS INEQUALITY. FOR EXAMPLE, LET’S TEST X=-2
ON THE LEFT. LETS TEST 0 IN THE MIDDLE,
AND LETS TEST 6 ON THE RIGHT. IF THESE VALUES SATISFY
THIS BLUE INEQUALITY, THEN THE ENTIRE INTERVAL IS
TRUE, AND THEREFORE THE DOMAIN. IF THE TEST VALUE IS FALSE, THEN THE ENTIRE INTERVAL
IS FALSE AND NOT IN THE DOMAIN
OF OUR FUNCTION. SO WHEN X IS -2, YOU WOULD HAVE
-2 SQUARED + 4 x -2 + 5. AND THE QUESTION IS,
IS THIS GREATER THAN OR=0. WELL, THIS WOULD BE -4 – 8 + 5
GREATER THAN OR=0. THIS IS GOING TO BE -7, SO IT’S
NOT GREATER THAN OR=0, THEREFORE THIS ENTIRE INTERVAL
FALSE. PUT AN F HERE TO REMIND US. NOW WE’LL TEST X=0, SO WE’D HAVE -0 SQUARED + 4 x 0
+ 5 GREATER THAN OR=0. THIS WOULD JUST BE 5,
WHICH IS GREATER THAN OR=0. THEREFORE,
THIS ENTIRE INTERVAL IS TRUE. SO I’LL GO AHEAD
AND GRAPH THIS INTERVAL. THIS INTERVAL IS IN THE DOMAIN
OF OUR FUNCTION. AND THE LAST TEST VALUE
IS X=6. SO WE’D HAVE -6 SQUARED + 4 x 6
+ 5 GREATER THAN OR=0. IT’S GOING TO BE -36 + 24
THAT’S -12 + 5 IS -7. WELL, THIS IS NOT GREATER THAN
OR EQUAL TO ZERO, AND THEREFORE THIS INTERVAL HERE
ON THE RIGHT IS FALSE. SO THE DOMAIN OF OUR FUNCTION
USING INTERVAL NOTATION WOULD BE THE CLOSE INTERVAL FROM
-1 TO 5, MEANING IT INCLUDES
THE END POINTS. BUT WE COULD ALSO SAY THAT X
IS GREATER THAN OR=-1 AND LESS THAN OR=5. NOW, THIS WAS QUITE
A BIT OF WORK, SO I DO WANT TO SHOW
HOW WE COULD’VE DETERMINED THIS DOMAIN A LOT FASTER
GRAPHICALLY. GOING BACK UP TO OUR RESTRICTION
OF THE DOMAIN OR THIS QUADRATIC INEQUALITY
HERE, WE COULD’VE GRAPHED THE FUNCTION
F OF X=-X SQUARED + 4X + 5 AND DETERMINED GRAPHICALLY WHEN THE FUNCTION WAS GREATER
THAN OR=0. SO LET’S GO AND TAKE A LOOK
AT THAT GRAPH NOW. SO IN RED WE HAVE THE GRAPH OF F
OF X=-X SQUARED + 4X + 5. TO DETERMINE WHICH X VALUE
SATISFY THIS INEQUALITY WE WANT TO KNOW
WHEN THIS FUNCTION IS GREATER THAN OR EQUAL
TO ZERO. WHICH MEANS ONE OF THE Y VALUES
GREATER THAN OR EQUAL TO ZERO, WHICH OCCURS WHEN THE FUNCTION
IS ABOVE THE X AXIS. NOTICE HOW THIS FUNCTION
IS ABOVE THE X AXIS ON THIS INTERVAL HERE. AND IF WE PROJECT
THIS ON TO THE X AXIS IT WOULD BE THE CLOSED INTERVAL
FROM -1 TO 5. AGAIN, WE’RE GOING TO INCLUDE
THE END POINTS BECAUSE WE WANT TO KNOW WHEN THIS IS GREATER THAN
OR EQUAL TO ZERO. AND BECAUSE OF THE EQUAL PART,
WE INCLUDE THE END POINTS. SO THIS GRAPH VERIFIES THAT WE FOUND THE DOMAIN BY HAND
CORRECTLY. SO THIS IS A NICE WAY
TO GRAPHICALLY SOLVE A QUADRATIC INEQUALITY. BUT OF COURSE THE OTHER WAY TO DETERMINE THE DOMAIN
OF A SQUARE ROOT FUNCTION WOULD BE TO JUST GRAPH
THE ORIGINAL FUNCTION, WHICH WE SEE HERE IN BLUE. AND WE CAN SEE FROM THE GRAPH
OF THE ORIGINAL FUNCTION, THE DOMAIN WOULD BE THE CLOSE
INTERVAL FROM -1 TO 5.   OKAY, I HOPE YOU FOUND THIS
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