- IF F OF X EQUALS X DIVIDED BY 
THE QUANTITY X + 3,
DETERMINE FOR WHAT X VALUES 
F PRIME OF X = 4.
SO BY DETERMINING FOR WHAT X 
VALUES F PRIME OF X = 4,
WE'LL BE DETERMINING 
WHERE ON THE FUNCTION
THE SLOPE OF THE TANGENT LINES 
WOULD BE FOUR.
AND SINCE F IS A QUOTIENT WE'LL 
HAVE TO APPLY THE QUOTIENT RULE
TO FIRST FIND THE DERIVATIVE 
FUNCTION.
THEN ONCE WE FOUND THE 
DERIVATIVE FUNCTION
WE CAN SET IT EQUAL TO FOUR 
AND SOLVE FOR X.
THE QUOTIENT RULE IS STATED HERE 
BELOW
WHERE THE DERIVATIVE OF FUNCTION 
F DIVIDED BY FUNCTION G
WITH RESPECT TO X IS EQUAL TO G 
x F PRIME - F x G PRIME
DIVIDED BY G SQUARED.
OR, IN OUR CASE, APPLYING THIS 
FORMULA F WOULD BE EQUAL TO X
AND G WOULD BE EQUAL TO X + 3.
SO F PRIME OF X WOULD BE EQUAL 
TO THE DENOMINATOR
WHICH IS THE QUANTITY X + 3 x 
THE DERIVATIVE OF THE NUMERATOR
WHICH WOULD BE THE DERIVATIVE 
OF X - THE NUMERATOR WHICH IS X
TIMES THE DERIVATIVE 
OF THE DENOMINATOR.
THE DERIVATIVE OF THE QUANTITY 
X = 3.
THIS IS DIVIDED 
BY THE (X + 3 SQUARED).
SO NOW WE'LL FIND THE DERIVATIVE 
HERE AND THE DERIVATIVE HERE.
AND NOW WE CAN SIMPLIFY.
SO WE WOULD HAVE (X + 3) 
x THE DERIVATIVE OF X,
WHICH IS JUST ONE.
SO WE JUST HAVE (X + 3)
AND THEN - X x THE DERIVATIVE OF 
X + 3 IS ALSO JUST ONE.
SO WE HAVE JUST X HERE DIVIDED 
BY THE QUANTITY X + 3 SQUARED.
WE'LL NOTICE HERE WE HAVE X - X.
SO THE X'S SIMPLIFY OUT,
AND THEREFORE OUR DERIVATIVE 
FUNCTION, F PRIME OF X = 3
DIVIDED BY (X + 3 SQUARED).
SO OUR GOAL IS TO DETERMINE 
FOR WHAT X VALUES
THE DERIVATIVE FUNCTION 
IS EQUAL TO FOUR.
SO NOW WE'LL SET THIS EQUAL 
TO FOUR AND SOLVE FOR X.
SO WE'D HAVE THREE DIVIDED BY 
(X + 3 SQUARED) = 4.
LET'S WRITE IT AS 4/1.
AND NOW TO SOLVE FOR X 
WE'RE CROSS MULTIPLY.
4 x (X + 3 SQUARED) 
MUST EQUAL 3 x 1.
SO 4 x (X + 3 SQUARED) 
MUST EQUAL 3 x 1 OR 3.
WE'RE NOT GOING TO EXPAND THIS.
WE'LL SOLVE THIS 
USING SQUARE ROOTS.
SO NOW WE'LL DIVIDE BOTH SIDES 
BY FOUR.
SO NOW WE HAVE (X + 3 SQUARED) 
= 3/4.
SO NOW WE'LL TAKE 
THE SQUARE ROOTS
OF BOTH SIDES OF THE EQUATION.
IN DOING THIS DON'T FORGET THE 
PLUS OR MINUS SIDE ON THE RIGHT
UNTIL WE OBTAIN BOTH SOLUTIONS.
SO NOW WE HAVE JUST X + 3 =--
THIS WOULD BE EQUAL TO PLUS OR 
MINUS THE SQUARE ROOT OF 3
OVER THE SQUARE ROOT OF FOUR.
BUT THE SQUARE ROOT OF 4 = 2.
SO WE HAVE X + 3 EQUALS PLUS 
OR MINUS SQUARE ROOT 3 OVER 2.
AND NOW WE'LL SUBTRACT THREE 
ON BOTH SIDES OF THE EQUATION
TO SOLVE FOR X.
WE WOULD HAVE X = -3 PLUS 
OR MINUS SQUARE ROOT 3
DIVIDED BY 2.
SO THE X VALUES 
FOR WHICH F PRIME OF X = 4
WOULD BE, SAY, X OF 1 = -3 
- SQUARE ROOT 3 DIVIDED BY 2.
AND THE OTHER X VALUE WOULD BE 
X OF 2 = -3 + SQUARE ROOT 3
DIVIDED BY 2.
WE ARE GOING TO SHOW 
WHAT THIS MEANS GRAPHICALLY
IN JUST A MOMENT.
BUT BY CHECKING IT GRAPHICALLY
IT WILL BE HELPFULLY TO HAVE 
THE DECIMAL APPROXIMATIONS
FOR THESE X VALUES.
SO THIS FIRST X VALUE 
IS APPROXIMATELY -3.866
AND THE SECOND X VALUE 
IS APPROXIMATELY -2.134.
NOW, WE'LL TAKE A LOOK 
AT THE GRAPH
AND THEN WE'LL FIND THE POINTS 
ON THE FUNCTION
WITH THESE X VALUES
AND THEN TAKE A LOOK 
AT THE TANGENT LINES
TO SEE IF THEY DO HAVE A SLOPE 
OF POSITIVE FOUR.
THIS POINT HERE IS WHERE X = -3 
+ SQUARE ROOT 3 DIVIDED BY 2.
AND THIS POINT HERE IS WHERE X = 
-3 - SQUARE ROOT 3 DIVIDED BY 2.
AND NOTICE HOW WHEN WE SKETCH 
THE TANGENT LINES,
AT THOSE POINTS THESE LINES 
DO HAVE A SLOPE OF +4
VERIFYING THAT OUR WORK 
IS CORRECT.
