1 Find the equation of the tangent line to the graph of the function at the indicated point.
a f (x) = x
2
; (1, 1) b f (x) = 4x
2
– 3x – 5; (−2, 17)
c f (x) = x
3
+ 1; (−1, 0) d
1
3
f x x ( ) ( 1) ; (1,0) = −
e f (x) = ( ) 1
; 1, 1
x
2 Let f (x) =
2
, 3
.
6, 3
x if x
x if x
>
− ≤
Find the equation of the line tangent to the graph
of f at each of the following points.
a (0, −6) b (−2, −2) c (1, −5) d (3, 3) e (4, 4)
3 Let f (x) = x
3
– 3x, find the values of x at which the slope of the tangent line is 0.
4 Let f (x) = x
3
+ x2
– x + 1; find the set of values of x such that the slope is positive.
5 If the graphs of the functions f and g given by
f (x) = 4 – x
2
and g (x) = x
3
– 8x have the same slope, determine the values of x.
6 Let f(x) = x
3
– x
2
– x + 1; find the equation of the tangent line at the point where the
graph crosses
a the x-axis b the y-axis c the graph of y = 1– x
2
�
1a) The function f(x) = x has a slope of 1 at every point. Therefore, the equation of the tangent line at (1,1) is y = x.
1b) The derivative of f(x) = 4x^2 - 3x - 5 is f'(x) = 8x - 3. Evaluating f'(-2) gives a slope of -19. Therefore, the equation of the tangent line at (-2,17) is y = -19x + 59.
1c) The derivative of f(x) = x^3 + 1 is f'(x) = 3x^2. Evaluating f'(-1) gives a slope of 3. Therefore, the equation of the tangent line at (-1,0) is y = 3(x+1).
1d) The derivative of f(x) = (1/3)x^3 - x + 1 is f'(x) = x^2 - 1. Evaluating f'(1) gives a slope of 0. Therefore, the equation of the tangent line at (1,0) is y = 0.
1e) The derivative of f(x) = (1/x) + 1 is f'(x) = -1/x^2. Evaluating f'(1) gives a slope of -1. Therefore, the equation of the tangent line at (1,1) is y = -x + 2.
2a) Since x ≤ 0 at (0,-6), we evaluate f(x) = x/2 at x = 0 and get f(0) = 0. The derivative of f(x) is f'(x) = 1/2 for x > 0, and f'(x) = 1 for x < 0. Since x = 0 is a corner point, we need to consider the left and right limits of the slope. The left limit is f'(0-) = 1, and the right limit is f'(0+) = 0. Therefore, the equation of the tangent line at (0,-6) is y = x + (-6).
2b) Since x > 0 at (-2,-2), we evaluate f(x) = 3/x at x = 2 and get f(2) = 3/2. The derivative of f(x) is f'(x) = -3/x^2 for x > 0. Therefore, the equation of the tangent line at (-2,-2) is y = (-3/4)x - (3/2).
2c) Since x > 0 at (1,-5), we evaluate f(x) = 3/x at x = 1 and get f(1) = 3. The derivative of f(x) is f'(x) = -3/x^2 for x > 0. Therefore, the equation of the tangent line at (1,-5) is y = -3(x-1) - 5.
2d) Since x ≤ 0 at (3,3), we evaluate f(x) = x/2 at x = 3 and get f(3) = 3/2. The derivative of f(x) is f'(x) = 1/2 for x > 0, and f'(x) = 1 for x < 0. Since x = 0 is a corner point, we need to consider the left and right limits of the slope. The left limit is f'(3-) = 1/2, and the right limit is f'(3+) = 0. Therefore, the equation of the tangent line at (3,3) is y = (1/2)(x-3) + 3.
2e) Since x > 0 at (4,4), we evaluate f(x) = 3/x at x = 4 and get f(4) = 3/4. The derivative of f(x) is f'(x) = -3/x^2 for x > 0. Therefore, the equation of the tangent line at (4,4) is y = -3(x-4)/16 + 4.
3 To find the values of x where the slope of the tangent line is 0, we need to set the derivative equal to 0 and solve for x:
f'(x) = 3x^2 - 3 = 0
3x^2 = 3
x^2 = 1
x = ±1
Therefore, the slope of the tangent line is 0 at x = -1 and x = 1.
4 To find the set of values of x such that the slope is positive, we need to find where the derivative is positive:
f'(x) = 3x^2 + 2x - 1 > 0
Using the quadratic formula to solve for the roots:
x = (-2 ± √10)/6
Therefore, the slope is positive when -2/3 < x < (√10 - 2)/6.
5 To find where the slopes of f(x) and g(x) are equal, we need to set their derivatives equal to each other:
f'(x) = -2x
g'(x) = 3x^2 - 8
-2x = 3x^2 - 8
3x^2 + 2x - 8 = 0
Solving for x using the quadratic formula:
x = (-2 ± √34)/6
Therefore, the slopes are equal at x = (-2 + √34)/6 and x = (-2 - √34)/6.
6a) To find where the graph of f crosses the x-axis, we need to solve for f(x) = 0:
x^3 - x^2 - x + 1 = 0
We can factor this as:
(x-1)(x^2-1) = 0
Therefore, the graph crosses the x-axis at x = 1. The equation of the tangent line at this point is y = f'(1)(x-1), where f'(x) is the derivative of f(x):
f'(x) = 3x^2 - 2x - 1
f'(1) = 0
Therefore, the equation of the tangent line at (1,0) is y = 0.
6b) To find where the graph of f crosses the y-axis, we need to set x = 0:
f(0) = 1
Therefore, the graph crosses the y-axis at (0,1). The equation of the tangent line at this point is y = f'(0)(x-0) + 1:
f'(x) = 3x^2 - 2x - 1
f'(0) = -1
Therefore, the equation of the tangent line at (0,1) is y = -x + 1.
6c) To find where the graph of f crosses the graph of y = 1 - x^2, we need to solve for f(x) = 1 - x^2:
x^3 - x^2 - x + 2 = 0
We can factor this as:
(x-2)(x^2+1) = 0
Therefore, the graph of f crosses the graph of y = 1 - x^2 at x = 2. The equation of the tangent line at this point is y = f'(2)(x-2) + 1 - 2:
f'(x) = 3x^2 - 2x - 1
f'(2) = 11
Therefore, the equation of the tangent line at (2,-3) is y = 11(x-2) - 3.