﻿ Mean Value Theorem Worksheet | Problems & Solutions Mean Value Theorem Worksheet

Mean Value Theorem Worksheet
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1.
We verify the "Mean Value theorem" for a function $f$ ($x$): a. If it is continuous on [$a$, $b$] b. If it is differentiable on ($a$, $b$) c. If it is continuous on ($a$ , $b$) d. Both A and B

Solution:

If f(x) is continuous on [a, b] and is differentiable on (a, b), then only it follows the hypotheses of mean value theorem for which the mean value theorem can be verified.

2.
State whether the function $f$ ($x$) = 3$x$2 - 2 on [2, 3] satisfies the mean value theorem. a. no b. yes

Solution:

f (x) = 3x2 - 2, x [2, 3]
[Given function.]

The given function is continuous on [2, 3] and differentiable in (2, 3).

So, there exists c (2, 3) such that f ′ (c) = f(3) - f(2)3 - 2.
[By mean value Theorem.]

f ′(c) = 15.
[f ′(c) = 6c. ]

6c = 15 c = 5 / 2 (2, 3).

Hence, the given function satisfies the mean value theorem.

3.
Which of the following functions satisfies the mean value theorem? a. $f$($x$) = $x$2 - 6 , $x$ $\in$ [7, 8] b. $f$ ($x$) = log $x$, $x$ $\in$ [-7, 8] c. $f$ ($x$) = [$x$], $x$ $\in$ [-7, 7] d. $f$ ($x$) = |$x$|, $x$ $\in$ [-8, 8]

Solution:

f (x) = x2 - 6, the function f (x) is continuous on [7, 8] and differentiable on (7, 8)
[Consider the choice A.]

f ′ (c ) = f (8) - f (7)8 - 7 = 15
[Try to get c (7, 8) such that f ′ (c ) = f (b) - f (a)b - a]

2c = 15
[f ′ (x) = 2x]

c = 15 / 2 [7, 8]
[There exists c (7, 8) such that f ′ (c ) = f (b) - f (a)b - a]

So, the function in the choices A satisfies the mean value theorem

f (x) = log x, x [- 7, 8] is not continuous in [- 7, 8] and hence does not satisfy the mean value theorem
[Consider the choice B.]

f (x) = [x], x [- 7, 7] is not continuous in [- 7, 7] and hence does not satisfy the mean value thoerem
[Consider the choice C.]

f (x) = |x|, x [- 8, 8] is continuous in [- 8, 8] but not differentiable in (- 8, 8) and hence does not satisfy the mean value theorem .
[Consider the choice D.]

4.
Determine the point on the parabola $f$ ($x$) = ($x$ - 2)2, at which the tangent is parallel to the chord joining the points (2, 0) & (3, 1). a. ($\frac{5}{2}$, $\frac{1}{4}$) b. ($\frac{3}{2}$, $\frac{1}{4}$) c. ($\frac{3}{2}$, $\frac{9}{4}$) d. ($\frac{3}{2}$, - $\frac{3}{2}$)

Solution:

Slope of the chord joining (2, 0) & (3, 1) = 1-0 / 3-2 = 1

Slope of the tangent to the curve at any point (x, f(x)) is f ′ (x) = 2(x - 2)

f ′ (c) = 1
[By Mean Value Theorem.]

2(c - 2) = 1

c = 5 / 2 (2, 3)

f (c) = (c - 2)2 = (5 / 2 - 2)2 = 1 / 4

Hence, the point where the tangent to the parabola is parallel to the given chord is (c, f (c)) = (5 / 2, 1 / 4)

5.
Find the point on the parabola $y$ = ($x$ + 3)2, at which the tangent is parallel to the chord of the parabola joining the points (- 3, 0) & (- 4, 1). a. (- 4, 1) b. (- $\frac{5}{2}$, $\frac{1}{4}$) c. (- 2, 1) d. (- $\frac{7}{2}$, $\frac{1}{4}$)

Solution:

Slope of the chord joining the points (- 3, 0) & (- 4, 1) = (1 -  0)( - 4 + 3) = - 1

Slope of the tangent to the curve at any point (x, f (x)) is f ′ (x ) = 2(x + 3)

f ′(c) = -1
[By mean value theorem.]

2(c + 3) = -1

c = - 7 / 2 (-4, - 3)

f (c) = (c + 3)2 = (- 72 + 3)2 = 14.

Hence, the point where the tangent to the parabola is parallel to the given chord is (c, f(c)) = (- 7 / 2, 1 / 4).

6.
Find the point on the graph of $f$ ($x$) = $x$3, at which the tangent is parallel to the chord joining the points (1, 1) & (3, 27). a. (${\left(\frac{13}{3}\right)}^{\frac{1}{2}}$, ${\left(\frac{13}{3}\right)}^{\frac{3}{2}}$) b. ($\frac{13}{3}$, $\frac{3}{13}$) c. (1, 1) d. (0, 0)

Solution:

Slope of the chord joining the points (1, 1) & (3, 27) is = 27 - 13 - 1 = 13

Slope of the tangent to the curve at any point (x, f(x)) is f ′(x) = 3x2.

f ′(c) = 13
[By mean value theorem.]

3 c2 = 13

c = (133)12 (1, 3).

f (c) = (133)32

Hence, the point where the tangent to the curve is parallel to the given chord is (c, f(c)) = ((133)12, (133)32)

7.
Find the point at which the tangent to the curve $f$($x$) = $x$2 - 6$x$ + 1 is parallel to the chord joining the points (1, - 4) & (3, -8). a. (2, -7) b. (2, 7) c. (-2, -7) d. (-2, 7)

Solution:

Slope of the chord joining the points (1, - 4) & (3, - 8) is - 8 + 43 - 1 = - 2

Slope of the tangent to the curve at any point (x, f(x)) is f ′(x) = 2x - 6

f ′(c) = - 2
[By Mean Value Theorem.]

2c - 6 = - 2

c = 2 (1, 3).

f(c) = (2)2 - 6(2) + 1

f(c) = - 7

Hence, the point where the tangent to the curve is parallel to the given chord is (c, f(c)) = (2, - 7)

8.
Find the point on the curve $y$ = 12($x$ + 1)($x$ - 2) in the interval [-1, 2], at which the tangent is parallel to the $x$ - axis. a. (- $\frac{1}{2}$, - 27) b. ($\frac{1}{2}$, - 27) c. (- $\frac{1}{2}$, 27) d. ($\frac{1}{2}$, 27)

Solution:

y = f(x) = 12(x + 1)(x - 2)
[Given curve.]

= 12x2 - 12x - 24

Slope of the tangent to the curve at any point (x, f (x)) is f ′(x) = 24x -12

f ′ (c) = f(2) - f(-1)2 + 1
[By Mean Value Theorem.]

24c - 12 = 0

c = 1 / 2 (-1, 2)

f(c) = 12(1 / 2)2 - 12(1 / 2) - 24

= 3 - 6 - 24

= - 27

Hence, the point where the tangent to the curve is parallel to x- axis is (c, f (c)) = (1 / 2, - 27).

9.
State whether the function $f$($x$) = ln $x$ on [1, 2] satisfies the mean value theorem. a. yes b. no

Solution:

f(x) = ln x, x [1, 2]
[Given function.]

f(x) is continuous on [1, 2] and differentiable on (1, 2).

So, there exists c (1, 2) such that f ′ (c) = f(2) - f(1)2 - 1.
[By mean value Theorem.]

1c = ln 2 c = 1ln 2 (1, 2).
[f ′(x) = 1x]

Hence, the given function satisfies the mean value theorem.

10.
State whether the function $f$ ($x$) = sin $x$ - sin 2$x$, $x$ $\in$ [0, $\pi$] satisfies the mean value theorem. a. yes b. no

Solution:

f (x) = sin x - sin 2x, x [0, π]
[Given function.]

The given function is continuous on [0, π] and differentiable in (0, π).

So, there exists c (0, π) such that f ′ (c) = f(π) - f(0)π  - 0.
[By mean value theorem.]

cos c - 2 cos 2 c = 0
[f ′ (x) = cos x - 2 cos 2x.]

cos c = 4 cos2c - 2
[cos 2c = 2 cos2c - 1.]

4 cos2 c - cos c - 2 = 0

c = 32o.53′ , 126o.37′ (0, π) )
[Solve use the calculator.]

Hence, the function f (x) satisfies the mean value theorem.