1 The Mean Value Theorem
1.1 Solution 1
\(f(x) = x^2 - 3x\), \(a=1\), \(b=4\)
\(f'(x) = 2x-3\)
By mean value thoerem,
\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)
\(2c-3 = \dfrac{4^2 - 3.4 - (1^2 - 3)}{4-1}\)
\(2c-3 = \dfrac{16-12-(-2)}{3} = \dfrac{4+2}{3} = \dfrac{6}{3}\)
\(2c-3 = 2\)
\(2c = 5\)
\(c = \dfrac{5}{2}\)
1.2 Solution 2
\(f(x) = x^3, a=-2, b=4\)
\(f'(x) = 3x^2\)
By mean value thoerem,
\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)
\(3c^2 = \dfrac{4^3 - (-2)^2}{4-(-2)} = \dfrac{4^3+2^3}{4+2}\)
\(3c^2 = \dfrac{72}{6} = 12\)
\(c^2 = 4\)
\(c = 2\)
1.3 Solution 3
\(f(x) = \sqrt{x}, a = 1, b=9\)
\(f'(x) = \dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}\)
By mean value thoerem,
\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)
\(\dfrac{1}{2\sqrt{c}} = \dfrac{f(9)-f(1)}{9-1} = \dfrac{\sqrt{9}-\sqrt{1}}{9-1}\)
\(\dfrac{1}{2\sqrt{c}} = \dfrac{3-1}{9-1} = \dfrac{2}{8} = \dfrac{1}{4}\)
\(\dfrac{1}{\sqrt{c}} = \dfrac{1}{2}\)
\(\sqrt{c} = 2\)
\(c = 4\)
1.4 Solution 4
\(f(x) = \sin x, a=0,b=\pi\)
\(f'(x) = \cos x\)
By mean value thoerem,
\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)
\(\cos c = \dfrac{\sin \pi - \sin 0}{\pi - 0} = \dfrac{0-0}{\pi}\)
\(\cos c = 0\)
\(c = \arccos 0 = \dfrac{\pi}{2}\)
1.5 Solution 5
\(f(x) = \dfrac{1}{x^2}, a=-4, b=-1\)
\(f(x) = x^{-2}\)
\(f'(x) = -2x^{-3}\)
By mean value thoerem,
\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)
\(-2c^{-3} = \dfrac{f(-1)-f(-4)}{-1-(-4)}\)
\(\dfrac{-2}{c^3} = \dfrac{1-(1/16)}{3} = \dfrac{15/16}{3} = \dfrac{5}{16}\)
\(\dfrac{-2}{c^3} = \dfrac{5}{16}\)
\(c^3 = \dfrac{-2.16}{5} = \dfrac{-32}{5}\)
\(c = - \sqrt[3]{\dfrac{32}{5}}\)
1.6 Solution 6
\(f(x) = x^{2/3}\)
\(-1 < c < 8\)
By mean value thoerem,
\(f'(c) = \dfrac{f(8)-f(-1)}{8 - (-1)}\)
Let's assume that such a number exits,
\(f'(c) = \dfrac{8^{2/3} - (-1)^{2/3}}{9} = \dfrac{2^2 - (-1)^2}{9}\)
\(= \dfrac{4-1}{9} = \dfrac{3}{9} = \dfrac{1}{3}\)
\(f'(c) = \dfrac{1}{3}\)
\(f(x) = x^{2/3}\)
\(f'(x) = \dfrac{2}{3}x^{2/3 - 1} = \dfrac{2}{3}x^{-1/3} = \dfrac{2}{3x^{1/3}}\)
\(f'(c) = \dfrac{2}{3c^{1/3}}\)
\(\dfrac{2}{3c^{1/3}} = \dfrac{1}{3}\)
\(c^{1/3} = 2\)
\(c = 8\)
But we know that \(-1 < c < 8\), so no such number exists.
It doesnt' contradict the mean value theorem because the function isn't differentiable at \(0\).
1.7 Solution 7
1.7.1 Solution a
\(\sqrt{10} < \dfrac{19}{6} \approx 3.167\)
Let \(f(x) = \sqrt{x}\)
\(f\) is continous and differentiable at all positive integer.
So, it is continous on \([9,10]\) and differentiable on \((9,10)\)
Mean value theorem tells us that,
\(\dfrac{\sqrt{10} - \sqrt{9}}{10 - 9} = f'(c)\)
\(f'(x) = \dfrac{1}{2} x^{-1/2} = \dfrac{1}{2\sqrt{x}}\)
\(\dfrac{1}{2\sqrt{c}} = \dfrac{\sqrt{10} - 3}{1} = \sqrt{10} - 3\)
\(\sqrt{10} = \dfrac{1}{2\sqrt{c}} + 3\)
We know that \(9 < c < 10\)
Since \(c>9\), we have \(\sqrt{c} > 3\).
\(1 > \dfrac{3}{\sqrt{c}}\)
\(\dfrac{1}{3} > \dfrac{1}{\sqrt{c}}\)
\(\dfrac{1}{6} > \dfrac{1}{2\sqrt{c}}\)
\(\dfrac{1}{2\sqrt{c}} < \dfrac{1}{6}\)
So, \(\dfrac{1}{2\sqrt{c}} + 3 < \dfrac{1}{6} + 3 = \dfrac{10}{6}\)
\(\dfrac{1}{2\sqrt{c}} + 3 < \dfrac{10}{6}\)
\(\sqrt{10} < \dfrac{10}{6}\)
1.7.2 Solution b
\(\sqrt{10} > \dfrac{60}{19} \approx 3.158\)
We know that \(c < 10\)
So, \(\sqrt{c} < \sqrt{10} < \dfrac{19}{6}\) (from a)
\(\sqrt{10} = \dfrac{1}{2\sqrt{c}} + 3 > 3 + \dfrac{1}{2.19/6} = 3 + \dfrac{3}{19}\)
\(\dfrac{1}{2\sqrt{c}} + 3 > \dfrac{60}{19}\)
\(\sqrt{10} > \dfrac{60}{19}\)
1.8 Solution 8
1.8.1 Solution a
\(\sqrt[3]{10} < \dfrac{13}{6} \approx 2.167\)
\(f(x) = \sqrt[3]{x}\)
\(f\) is continous on all real numbers and differentiable too.
Interval = \([8,10]\)
\(f'(x) = \dfrac{1}{3}x^{1/3 - 1} = \dfrac{1}{3}x^{-2/3} = \dfrac{1}{3x^{2/3}}\)
\(f'(c) = \dfrac{f(10)-f(8)}{10-8}\)
\(\dfrac{1}{3c^{2/3}} = \dfrac{10^{1/3} - 8^{1/3}}{2} = \dfrac{10^{1/3}-2}{2}\)
\(\dfrac{1}{3c^{2/3}} = \dfrac{10^{1/3}-2}{2}\)
\(3c^{2/3} = \dfrac{2}{10^{1/3}-2}\)
\(10^{1/3} - 2 = \dfrac{2}{3c^{2/3}}\)
\(\sqrt[3]{10} = \dfrac{2}{3c^{2/3}} + 2\)
We know that \(8 < c < 10\)
Since \(c > 8\), we have \(\sqrt[3]{c} > 2\)
\(c^{2/3} > 4\)
\(3c^{2/3} > 12\)
\(\dfrac{1}{12} > \dfrac{1}{3c^{2/3}}\)
\(\dfrac{1}{6} > \dfrac{2}{3c^{2/3}}\)
\(\dfrac{1}{6} +2 > \dfrac{2}{3c^{2/3}} + 2\)
\(\dfrac{13}{6} > \dfrac{2}{3c^{2/3}} + 2\)
\(\dfrac{13}{6} > \sqrt[3]{10}\)
\(3\sqrt{10} < \dfrac{13}{6}\)
1.8.2 Solution b
\(\sqrt[3]{10} > \dfrac{362}{169} \approx 2.142\)
We know that \(c < 10\)
So, \(\sqrt[3]{c} < \sqrt[3]{10} < \dfrac{13}{6}\) (from (a))
\(c^{1/3} < \dfrac{13}{6}\)
\(c^{2/3} < \dfrac{169}{36}>\)
\(3c^{2/3} < \dfrac{169}{12}\)
\(\sqrt[3]{10} = \dfrac{2}{3c^{2/3}} + 2 > \dfrac{2.12}{169} + 2 = \dfrac{362}{169}\)
\(\sqrt[3]{10} > \dfrac{362}{169}\)
1.9 Solution 9
\(\forall x x > 0 f'(x) = \dfrac{1}{x}\)
\(f(1) = 0\)
1.9.1 Solution a
\(\dfrac{1}{2} < f(2) < 1\)
Applying mean value theorem on the interval \([1,2]\).
Since \(f(x)\) is differentiable on all positive number, it's also continous. So \(f(x)\) is differentiable and continous on \([1,2]\)
\(f'(c) = \dfrac{f(2) - f(1)}{2-1}\)
\(\dfrac{1}{c} = \dfrac{f(2)-0}{1} = \dfrac{f(2)}{1}\)
\(\dfrac{1}{c} = f(2)\)
We know that \(1 < c < 2\)
So, \(\dfrac{1}{c} < 1\)
\(f(2) < 1\)
We know that \(c < 2\)
\(\dfrac{1}{2} < \dfrac{1}{c}\)
\(\dfrac{1}{2} < f(2)\)
Combinining them,
\(\dfrac{1}{2} < f(2) < 1\)
1.9.2 Solution b
\(\dfrac{1}{3} < f(3/2) < \dfrac{1}{2}\)
Applying mean value theorem on the interval \([1,3/2]\).
\(f'(c) = \dfrac{f(3/2)-f(1)}{3/2 - 1}\)
\(f'(c) = \dfrac{f(1.5)-0}{0.5} = \dfrac{f(1.5)}{0.5}\)
\(\dfrac{1}{c} = \dfrac{f(1.5)}{0.5}\)
\(f(1.5) = \dfrac{0.5}{c} = \dfrac{1}{2c}\)
We know that \(1 < c < 3/2\)
So, \(\dfrac{1}{2c} < \dfrac{1}{2}\)
\(f(3/2) < \dfrac{1}{2}\)
We know that \(c < 3/2\)
\(\dfrac{c}{2} < \dfrac{3}{4}\)
\(2 < \dfrac{3}{c}\)
\(1 < \dfrac{3}{2c}\)
\(\dfrac{1}{3} < \dfrac{1}{2c}\)
\(\dfrac{1}{3} < f(3/2)\)
Combinining both, \(\dfrac{1}{3} < f(3/2) < \dfrac{1}{2}\)
1.9.3 Solution c
\(\dfrac{7}{12} < f(2) < \dfrac{5}{6}\)
Applying mean value theorem on the interval \([3/2,2]\).
\(f'(c) = \dfrac{f(2) - f(1.5)}{2-1.5}\)
\(f'(c) = \dfrac{f(2)-f(1.5)}{0.5}\)
\(f'(c) = 2(f(2)-f(1.5))\)
\(\dfrac{1}{c} = 2(f(2) - f(1.5))\)
\(\dfrac{1}{2c} = f(2) - f(1.5)\)
\(f(2) = \dfrac{1}{2c} + f(1.5)\)
We know that \(\dfrac{3}{2} < c < 2\)
\(\dfrac{1}{2c} < \dfrac{1}{3}\)
We also know that \(\dfrac{1}{3} < f(1.5) < \dfrac{1}{2}\)
\(\dfrac{1}{2c} + f(1.5) < \dfrac{1}{3} + f(1.5)\)
\(f(2) < \dfrac{1}{3} + f(1.5) < \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{5}{6}\)
\(f(2) < \dfrac{5}{6}\)
We also know that, \(c < 2\)
So, \(\dfrac{1}{2} < \dfrac{1}{c}\)
\(\dfrac{1}{4} < \dfrac{1}{2c}\)
\(\dfrac{1}{4} + f(1.5) < \dfrac{1}{2c} + f(1.5)\)
\(\dfrac{1}{4} + f(1.5) < f(2)\)
Since \(\dfrac{1}{3} < f(1.5)\)
\(\dfrac{1}{4} + \dfrac{1}{3} < \dfrac{1}{4} + f(1.5) < f(2)\)
\(\dfrac{7}{12} < f(2)\)
Combinining both,
\(\dfrac{7}{12} < f(2) < \dfrac{5}{6}\)
1.10 Solution 10
Trick to this question is solving the inequality and seeing where it leads. Solving it leads to \(0 < b^2 + 9b\)
The easier way to solve it seems to be without using mean value theorem! (If anybody figures out to solve it via that, patches welcome!)
Suppose \(b > 0\)
\(b^2 > 0\)
\(b^2 + 9b > 0\)
\(b^2 + 9b + 27 > 27\)
\(b^3 + 9b^2 + 27b > 27b\)
\(b^3 + 3^3 + 3b3^2 + 3b^23^2 > 27b + 27\)
\((b+3)^3 > 27(b+1)\)
\(27(b+1) < (b+3)^3\)
\((b+1) < \dfrac{(b+3)^3}{3^3}\)
\(\sqrt[3]{b+1} < \dfrac{b+3}{3}\)
\(\sqrt[3]{b+1} < \dfrac{b}{3} + 1\)
1.11 Solution 11
\(\forall x > 0 f'(x) = \sqrt{x^2 + 4}\)
\(f(0) = 5\)
We need to prove \(\forall x > 0 f(x) > 2x+5\)
\(f'(x)\) is differentiable for all \(x > 0\). So it is also continous.
Let \(a\) b e an arbitrary element \(> 0\). Applying mean value theorem on \([0,a]\).
\(\dfrac{f(a)-f(0)}{a} = f'(c)\)
\(f(a) - 5 = a(f'(c))\)
\(f(a) = af'(c) + 5\)
\(f(a) = a\sqrt{c^2 + 4} + 5\)
We know that \(0 < c < a\)
\(0 < c^2 < a^2\)
\(4 < c^2 + 4 < a^2 + 4\)
\(2 < \sqrt{c^2 + 4} < \sqrt{a^2 + 4}\)
\(2a < a\sqrt{c^2 + 4} < a\sqrt{a^2 + 4}\)
\(2a + 5 < a\sqrt{c^2 + 4} + 5 < a\sqrt{a^2 + 4} + 5\)
So, \(2a + 5 < f(a)\)
Since \(a\) is arbitrary,
\(\forall x > 0, f(x) > 2x + 5\)
1.12 Solution 12
1.12.1 Solution a
Let \(f(x) = \sin x\)
From theorem 2.7.11, all trignometric functions are continous. Also, \(\sin x\) is differentiable for \(x > 0\).
If \(x > 2\pi\),
\(\sin x \leq 1 < 2\pi < x\)
So, \(\sin x < x\)
If \(0 < x \leq 2\pi\)
Applying mean value theorem on \([0,x]\),
\(\dfrac{\sin x - \sin 0}{x-0} = \cos c\)
\(\sin x = x\cos c\)
\(\cos c = \dfrac{\sin x}{x}\)
We know that \(0 < c < x\)
So, \(0 < c \leq 2\pi\)
So, \(\cos c < 1\)
So, \(\dfrac{\sin x}{x} < 1\)
\(\sin x < 1\)
Hence proved.
1.12.2 Solution b
From (a), we know that \(x > 0 \implies \sin x < x\)
From (a) we also know that \(0 < c < x\) and \(\cos c = \dfrac{\sin x}{x}\)
Suppose \(0 < x \leq \pi/2\)
So, \(0 < c < x \leq \pi/2\)
For the above range, \(\cos c > \cos x\)
\(\dfrac{\sin x}{x} > \sqrt{1 - \sin^2 x}\)
\(\dfrac{\sin^2 x}{x^2} > 1 - \sin^2 x\)
\(\sin^2 x > x^2 - x^2\sin^2 x\)
\(\sin^2 x(1 + x^2) > x^2\)
\(\sin^2 x > \dfrac{x^2}{1+x^2}\)
\(\sin x > \dfrac{x}{\sqrt{x^2 + 1}}\)
1.12.3 Solution c
From (a),
\(\sin (1/2) < \dfrac{1}{2}\)
From (b),
\(\sin(1/2) > \dfrac{1/2}{\sqrt{1/4+1}}\)
\(\sin(1/2) > \dfrac{1/2}{\sqrt{5/4}}\)
\(\sin(1/2) > \dfrac{1/2 * 2}{\sqrt{5}}\)
\(\sin(1/2) > \dfrac{1}{\sqrt{5}}\)
Combinining both,
\(\dfrac{1}{\sqrt{5}} < \sin (1/2) < 1/2\)
1.13 Solution 13
\(0 < x < \dfrac{\pi}{2} \implies \tan x > x\)
Let \(f(x) = \tan x\)
\(f(x)\) is continous on \([0,x]\) and differentiable on \((0,x/2)\)
Applying mean value theorem on \([0,x]\)
\(f'(c) = \dfrac{\tan x - \tan 0}{x}\)
\(\sec^2 x = \dfrac{\tan x}{x}\)
\(0 < c < x\)
For \(0 < c < x < \pi/2\)
\(\sec^2 0 < \sec^2 c < \sec^2 x\)
\(1 < \dfrac{\tan x}{x} < \sec^2 x\)
\(x < \tan x\)
\(\tan x > x\)
1.14 Solution 14
\(0 < \alpha < \pi/2\)
\(\sin \alpha = \dfrac{3}{5}\)
We need to prove \(\dfrac{3}{5} < \alpha < \dfrac{3}{4}\)
\(f(x) = \sin x\)
\(f(x)\) is continous for all the values and is differentiable.
Apply mean value theorem on \([0, \alpha]\)
\(\dfrac{f(\alpha) - f(0)}{\alpha} = f'(c)\)
\(\dfrac{\sin \alpha}{\alpha} = \cos c\)
We know that \(0 < c < \alpha < \dfrac{\pi}{2}\)
So, \(\cos \alpha < \cos c < 1\)
\(\sin \alpha = \dfrac{3}{5}\)
\(\dfrac{3/5}{\alpha} = \cos c\)
\(\dfrac{3}{5\alpha} = \cos c\)
\(\cos \alpha < \dfrac{3}{5\alpha} < 1\)
\(\cos \alpha = \sqrt{1-\sin^2 \alpha} = \sqrt{1 - 9/25} = \sqrt{16/25} = \dfrac{4}{5}\)
\(\dfrac{4}{5} < \dfrac{3}{5\alpha} < 1\)
\(\dfrac{4}{5}\alpha < \dfrac{3}{5} < \alpha\)
So, \(\dfrac{3}{5} < \alpha\)
\(\dfrac{4}{5}\alpha < \dfrac{3}{5}\)
\(4\alpha < 3\)
\(\alpha < \dfrac{3}{4}\)
Combinining both,
\(\dfrac{3}{5} < \alpha < \dfrac{3}{4}\)
1.15 Solution 15
\(f\) is continous on \([u,v]\)
We need to prove \(g\) is continous on \([p,q]\)
\(d = \dfrac{v-u}{3}\)
\(p = u + d\)
\(q = u + 2d\)
\(g(x) = \dfrac{f(x+d) - f(x)}{d}\)
where \(x\) is an arbitrary element in \([p,q]\). The maximum value of \(x+d\) is $q+d$and that is equal to \(v\).
So, \(f(x+d)\) is continous.
\(f(x)\) is continous since \([p,q]\) is a subset within \([u,v]\)
From theorem 2.7.3,
\(f(x+d)-f(x)\) is continous.
\(d = \dfrac{v-u}{3}\). So \(d\) is a number between \([v,u]\). And since \(u < v\), we know that \(d \neq 0\). So, \(\dfrac{f(x+d)-f(x)}{d}\) is continous from theorem 2.7.3 since \(d \neq 0\)