- 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.