- WE WANT TO DETERMINE THE SLOPE OF THE TANGENT LINES
AT X = 0 AND X = 1.
SO TO DO THIS,
WE'LL HAVE TO DETERMINE THE DERIVATIVE FUNCTION
AND THEN EVALUATE IT AT X = 0 AND X = 1.
AND SINCE THE GIVEN FUNCTION IS A QUOTIENT
OF TWO DIFFERENTIABLE FUNCTIONS,
WE'LL NEED TO APPLY THE QUOTIENT RULE TO FIND THE DERIVATIVE,
WHICH IS GIVEN HERE IN RED FOR REFERENCE.
SO WE'LL LET THE FUNCTION IN THE NUMERATOR EQUAL F
AND THE FUNCTION IN THE DENOMINATOR EQUAL G.
SO TO DETERMINE F PRIME OF X,
WE'LL START WITH THE DENOMINATOR,
WHICH IS JUST G SQUARED OR THE DENOMINATOR SQUARED.
SO WE'LL HAVE THE QUANTITY X SQUARED + 1 SQUARED,
AND FOR OUR NUMERATOR WE'LL HAVE G x F PRIME - F x G PRIME.
WELL, THE DENOMINATOR x THE DERIVATIVE OF THE NUMERATOR,
WHICH WOULD BE -2 - THE NUMERATOR, WHICH IS -2 X
x THE DERIVATIVE OF THE DENOMINATOR,
WHICH WOULD BE 2X.
SO NOW, WE'LL GO AHEAD AND SIMPLIFY THIS
AND THEN EVALUATE THIS DERIVATIVE FUNCTION
AT ZERO AND ONE.
SO HERE WE'LL DISTRIBUTE A -2.
SO WE'LL HAVE -2X SQUARED - 2
AND -2X x 2X = -4X SQUARED,
BUT SINCE WE'RE SUBTRACTING -4X SQUARED,
IT BECOMES + 4X SQUARED.
NOW, WE'LL COMBINE THE LIKE TERMS IN THE NUMERATOR.
SO OUR DENOMINATOR STAYS THE SAME.
SO WE'LL HAVE +2X SQUARED - 2,
AND THERE'S A COMMON FACTOR OF 2, AND THEN THIS DOES FACTOR.
IF WE FACTOR A TWO OUT, WE'LL HAVE X SQUARED - 1,
AND THEN HERE WE HAVE A DIFFERENCE OF SQUARES.
SO WE HAVE 2 x X + 1 x X - 1
ALL OVER THE QUANTITY X SQUARED + 1 SQUARED.
SO HERE'S OUR DERIVATIVE FUNCTION,
BUT OUR GOAL IS TO DETERMINE THE SLOPES OF THE TANGENT LINES
AT X = 0 AND X = 1.
SO FOR F PRIME OF 0
IT'D PROBABLY BE EASIER TO EVALUATE THE DERIVATIVE
IN THIS FORM HERE.
IF X IS ZERO, OUR NUMERATOR IS -2,
AND OUR DENOMINATOR WOULD BE ONE SQUARED OR ONE.
SO THE SLOPE OF THE TANGENT LINE AT X = 0 SHOULD BE -2.
WE'LL CHECK THAT IN JUST A MINUTE.
WE ALSO WANT TO KNOW THE SLOPE OF THE TANGENT LINE AT X = 1.
SO NOW, WE'LL EVALUATE THE DERIVATIVE AT X = 1,
AND AGAIN, IT DOESN'T MATTER WHICH OF THESE FORMS WE USE
TO EVALUATE THIS DERIVATIVE,
BUT LOOKING AT THE FACTORED FORM OF OUR DERIVATIVE,
NOTICE THAT IF X IS EQUAL TO ONE,
WE HAVE A FACTOR OF ZERO IN THE NUMERATOR,
AND OUR DENOMINATOR WOULD BE 1 + 2 SQUARED OR 4.
0/4 = 0,
SO AT X = 0
WE SHOULD HAVE A TANGENT LINE WITH SLOPE OF -2,
AND AT X = 1 WE SHOULD HAVE A TANGENT LINE
WITH A SLOPE OF ZERO.
LET'S GO AHEAD AND CHECK THAT GRAPHICALLY.
NOTICE AT X = 0 THE POINT IS ON THE ORIGIN,
AND THIS BLUE TANGENT LINE DOES HAVE A SLOPE OF -2.
NOTICE FROM THIS POINT TO THIS POINT
WE HAVE TO GO DOWN TWO UNITS AND RIGHT ONE,
AND WHEN X = 1, THAT WOULD BE THIS POINT HERE, (1, -1),
AND WE HAVE A HORIZONTAL TANGENT LINE,
WHICH MEANS THE SLOPE OF THIS TANGENT LINE WOULD BE ZERO.
SO THIS DOES VERIFY OUR WORK.
I HOPE THIS EXAMPLE WAS HELPFUL.