If the line 4x-9_y = 0 is tangent in the first quadrant to the graph of y = \x + c, what is the value of c
9
y' = x
2
If we rewrite the equation 4x-9y = 0 as y = l,x, we see that the slope of the line is §
Hence, the slope of the tangent line is 5, which must equal the derivative x
2
So, x = ± I
Since the point of
tangency is in the first quadrant, x = 3 The corresponding point on the line has y-coordmate \ = gx =
= 4 Since this point of tangency is also on the curve y = \x^ + c, we have ^ = j(^)
3 + c So,
c=JS?
For what nonnegative value(s) of b is the line y = - -fax + b normal to the graph of y = x
3 + 3
9
\' = 3x
2
Since the slope of y = -jjX + b is-^> the slope of the tangent line at the point of
intersection with the curve is the negative reciprocal of — n, namely 12 This slope is equal to the derivative
3x
2
Hence, x
2 = 4, and x = ±2 The ^-coordinate at the point of intersection is y = x^ + \ = (±2)
3 +
1 = T or -T So, the possible points are (2, J) and (-2,-T) Substituting in y = - ,2*-r f>, we
obtain b — *} and b = — ^ Thus, b = -y is the only nonnegative value
A certain point (x 0 , y 0 ) is on the graph of y = x^ + x
2 — 9\ — 9, and the tangent line to the graph at (x n , _y fl )
passes through the point (4, -1)
Find (x a , y (l )
> ' = 3x
2 + 2x - 9 is the slope of the tangent line This slope is also equal to (y + !)/(* - 4) Hence,
y + 1 = (3x
2 + 2\ — 9)(x - 4) Multiplying out and simplifying, y = 3*
1 - 10x
2 - \lx + 35 But the equation
y = x* + x
2 — 9x - 9 is also satisfied at the point of tangency Hence, 3*
1 - IQx
2 - 11 x + 35 = x
3 + A
: - 9x —
9 Simplifying, 2x — HA~ — 8x+ 44 = 0 In searching for roots of this equation, we first try integral factors of
44 It turns out that A =2 is a root So, A-2 is a factor of 2A
1 - 11*' - 8x + 44 Dividing 2*
3 -
llA2 -8vr + 44 by x-2, we obtain 2x -lx-22, which factors into (2x - H)(A -t 2) Hence, the
solutions are A = 2, x =—2, and x = V The corresponding points are (2, -15), (-2, 5), and (4
1
4
as
)
Answer
Let / be differentiate (rhat is, /' exists)
Define a function /* by the equation
/*(*) =
THE DERIVATIVE
51
evaluated, which is, therefore, equal to/'(3)• But, f'(x) = 20* . So, the value of the limit is 20(j)
3 =f^.
Answer
8.18
8.19
8.20
8.21
8.22
8.23
Thus
But
A function /, defined for all real numbers, is such that (/) /(I) = 2, (//) /(2) = 8, and (Hi) f(u + v)f(u) = kuv - 2v
2
for all u and u, where k is some constant. Find f'(x) for arbitrary x.
Substituting u = 1 and v = l in (Hi) and using (/) and (//), we find that k = 8. Now, in (/'//), let
and
Then
So,
Thus,
Find the points on the curve
where the tangent line is parallel to the line y = 3x.
y' = x
2 -l is the slope of the tangent line. To be parallel to the line y=3x having slope 3, it also must
have slope 3. Hence, x
2 - 1 = 3, *
2 = 4, x = ±2. Thus, the points are (2, f) and (-2, - j).
Recall the definition of the derivative: When
If we replace AJC by h in this limit, we obtain the limit to be
Find the relationship between /* and /'.
In particular,
for
f(x)=5x4,
Where
u=x
9
y' = x
2
If we rewrite the equation 4x-9y = 0 as y = l,x, we see that the slope of the line is §
Hence, the slope of the tangent line is 5, which must equal the derivative x
2
So, x = ± I
Since the point of
tangency is in the first quadrant, x = 3 The corresponding point on the line has y-coordmate \ = gx =
= 4 Since this point of tangency is also on the curve y = \x^ + c, we have ^ = j(^)
3 + c So,
c=JS?
For what nonnegative value(s) of b is the line y = - -fax + b normal to the graph of y = x
3 + 3
9
\' = 3x
2
Since the slope of y = -jjX + b is-^> the slope of the tangent line at the point of
intersection with the curve is the negative reciprocal of — n, namely 12 This slope is equal to the derivative
3x
2
Hence, x
2 = 4, and x = ±2 The ^-coordinate at the point of intersection is y = x^ + \ = (±2)
3 +
1 = T or -T So, the possible points are (2, J) and (-2,-T) Substituting in y = - ,2*-r f>, we
obtain b — *} and b = — ^ Thus, b = -y is the only nonnegative value
A certain point (x 0 , y 0 ) is on the graph of y = x^ + x
2 — 9\ — 9, and the tangent line to the graph at (x n , _y fl )
passes through the point (4, -1)
Find (x a , y (l )
> ' = 3x
2 + 2x - 9 is the slope of the tangent line This slope is also equal to (y + !)/(* - 4) Hence,
y + 1 = (3x
2 + 2\ — 9)(x - 4) Multiplying out and simplifying, y = 3*
1 - 10x
2 - \lx + 35 But the equation
y = x* + x
2 — 9x - 9 is also satisfied at the point of tangency Hence, 3*
1 - IQx
2 - 11 x + 35 = x
3 + A
: - 9x —
9 Simplifying, 2x — HA~ — 8x+ 44 = 0 In searching for roots of this equation, we first try integral factors of
44 It turns out that A =2 is a root So, A-2 is a factor of 2A
1 - 11*' - 8x + 44 Dividing 2*
3 -
llA2 -8vr + 44 by x-2, we obtain 2x -lx-22, which factors into (2x - H)(A -t 2) Hence, the
solutions are A = 2, x =—2, and x = V The corresponding points are (2, -15), (-2, 5), and (4
1
4
as
)
Answer
Let / be differentiate (rhat is, /' exists)
Define a function /* by the equation
/*(*) =
THE DERIVATIVE
51
evaluated, which is, therefore, equal to/'(3)• But, f'(x) = 20* . So, the value of the limit is 20(j)
3 =f^.
Answer
8.18
8.19
8.20
8.21
8.22
8.23
Thus
But
A function /, defined for all real numbers, is such that (/) /(I) = 2, (//) /(2) = 8, and (Hi) f(u + v)f(u) = kuv - 2v
2
for all u and u, where k is some constant. Find f'(x) for arbitrary x.
Substituting u = 1 and v = l in (Hi) and using (/) and (//), we find that k = 8. Now, in (/'//), let
and
Then
So,
Thus,
Find the points on the curve
where the tangent line is parallel to the line y = 3x.
y' = x
2 -l is the slope of the tangent line. To be parallel to the line y=3x having slope 3, it also must
have slope 3. Hence, x
2 - 1 = 3, *
2 = 4, x = ±2. Thus, the points are (2, f) and (-2, - j).
Recall the definition of the derivative: When
If we replace AJC by h in this limit, we obtain the limit to be
Find the relationship between /* and /'.
In particular,
for
f(x)=5x4,
Where
u=x