Calculus I
Notes on Intermediate Value Theorem.
The Intermediate Value Theorem: The Intermediate Value Theorem
If 1) a, b, c and k are real numbers,
2) f(x) is continuous on [a,b],
and 3) k is between f(a) and f(b),
then there is at least one value of c in (a,b) such that
f (x) = k .
The following graph demonstrates this theorem.
Basically, since f(x) is continuous, it can’t move from below the line y = k to above the line or visa-versa without crossing the line.
Example #1: For f (x) = x 3 − 3x 2 − 4x + 17 , find all values of c in [- 4, 0] such that f (c) = 5 .
►
First we need to confirm that the IVT applies.
f(- 4) = (- 4)3 − 3(- 4) 2 − 4(- 4) + 17 = -79
f(0) = (0)3 − 3(0) 2 − 4(0) + 17 = 17
and -79 ≤ 5 ≤ 17
Also f(x) is continuous. Thus, the IVT applies.
Now setup and solve the equation.
f (x) = k
x − 3x − 4x + 17 = 5
3
2
x 3 − 3x 2 − 4x + 12 = 0
x 2 ( x − 3) − 4 ( x − 3) = 0
( x − 3) ( x 2 − 4 ) = 0
( x − 3)( x − 2 )( x + 2 ) = 0
x = 3, 2, -2
Finally, we state the values of c remembering the interval.
c = -2.□
SCC:Rickman
Notes on Intermediate Value Theorem.
Page #1 of 2
⎛x⎞
Example #2: For f (x) = sin ⎜ ⎟ , approximate to the 3rd decimal place all values of c in
⎝2⎠
►
⎛ π⎞
⎜- ⎟
1
⎡ π⎤
sin ⎜ 3 ⎟ = sin ⎢ - ⎥ = - = -0.5
2
⎣ 6⎦
⎜⎜ 2 ⎟⎟
⎝
⎠
⎛π⎞
⎜ ⎟
2
⎡π⎤
sin ⎜ 2 ⎟ = sin ⎢ ⎥ =
≈ 0.707
2
4
2
⎣
⎦
⎜⎜ ⎟⎟
⎝ ⎠
and -0.5 ≤ -0.4 ≤ 0.707
⎡ π π⎤
⎢ - 3 , 2 ⎥ such that f (c) = -0.4 .
⎣
⎦
⎛x⎞
sin ⎜ ⎟ = -0.4
⎝2⎠
x
= sin −1 (-0.4)
2
x = 2sin −1 (-0.4)
x ≈ -0.823
c ≈ -0.823
Note, that we can simply use the sin (x) function because of the range of values for c.□
−1
Example #3: Approximate 2 to the 2nd decimal place.
►
This is one uses of the IVT. First, since 12 = 1, 22 = 4, and 1 < 2 < 4, we know that 1 < 2 < 2 . So find the midpoint of the interval
[1,2] which is 1.5. Square it. 1.52=2.25 which is greater than 2. Thus, 1 < 2 < 1.5 .
We now repeat this process until the endpoints round to the same number. The work above is in the table.
Interval
Midpoint
Midpoint2
> or < 2 New Interval
[1,2]
1.5
2.25
>
[1,1.5]≈ [1.00,1.50]
[1,1.5]
1.25
1.5625
<
[1.25,1.5] ≈[1.25,1.50]
[1.25,1.5]
1.375
1.890625
<
[1.375,1.5] ≈[1.38,1.50]
[1.375,1.5]
1.4675
2.06640625
>
[1.375,1.4675] ≈[1.38,1.47]
[1.375,1.4675]
1.40625
1.9775390625
<
[1.40625,1.4675] ≈[1.41,1.47]
[1.40625,1.4675]
1.421875
2.02172851563 >
[1.40625,1.421875] ≈[1.41,1.42]
[1.40625,1.421875]
1.4140625
1.99957275391 <
[1.4140625,1.421875] ≈[1.41,1.42]
[1.4140625,1.421875]
1.41796875
2.01063537598 >
[1.4140625,1.41796875] ≈[1.41,1.42]
[1.4140625,1.41796875]
1.416015625
2.00510025024 >
[1.4140625, 1.416015625] ≈[1.41,1.42]
[1.4140625, 1.416015625]
1.4150390625
2.0023355484
>
[1.4140625, 1.4150390625] ≈[1.41,1.42]
[1.4140625, 1.4150390625] 1.41455078125 2.00095391274 >
[1.4140625, 1.41455078125] ≈[1.41,1.41]
Thus, since 2 ∈ [1.4140625, 1.41455078125], 2 ≈ 1.41 .□
True, the method is example #3 is a lot of calculation, but if you didn’t have a calculator, a table of square roots, or calculus tools like
the derivative, this would be 1 method to generate a table of square roots or program a calculator to find square roots. We will get a
better method in Calculus I that uses the derivative.
SCC:Rickman
Notes on Intermediate Value Theorem.
Page #2 of 2