Infi-1 11

1a

\begin{matrix} Differntiate by definition: f (x) = x^{3} - 5 x \\ Solution: \\ f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = lim_{h \to 0} \frac{(x + h)^{3} - 5 (x + h) - (x^{3} - 5 x)}{h} = \\ = lim_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} - 5 x - 5 h - x^{3} + 5 x}{h} = lim_{h \to 0} 3 x^{2} + 3 x h + h^{2} - 5 = 3 x^{2} - 5 \\ ⟹ f^{'} (x) = 3 x^{2} - 5 \end{matrix}

1b

\begin{matrix} Differentiate by definition: f (x) = \sqrt{x^{4} + 1} \\ Solution: \\ f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = lim_{h \to 0} \frac{\sqrt{(x + h)^{4} + 1} - \sqrt{x^{4} + 1}}{h} = \\ = lim_{h \to 0} \frac{(x + h)^{4} + 1 - (x^{4} + 1)}{h (\sqrt{(x + h)^{4} + 1} + \sqrt{x^{4} + 1})} = lim_{h \to 0} \frac{((x + h)^{2} - x^{2}) ((x + h)^{2} + x^{2})}{h (\sqrt{(x + h)^{4} + 1} + \sqrt{x^{4} + 1})} = \\ = lim_{h \to 0} \frac{(x + h - x) (x + h + x) (x^{2} + 2 x h + h^{2} + x^{2})}{h (\sqrt{(x + h)^{4} + 1} + \sqrt{x^{4} + 1})} = \\ = lim_{h \to 0} \frac{(\overset{\to 2 x}{\overset{⏞}{2 x + h}}) (\overset{\to 2 x^{2}}{\overset{⏞}{2 x^{2} + 2 x h + h^{2}}})}{\underset{\to \sqrt{x^{4} + 1}}{\underset{⏟}{\sqrt{(x + h)^{4} + 1}}} + \sqrt{x^{4} + 1}} = \frac{2 x \cdot 2 x^{2}}{2 \sqrt{x^{4} + 1}} = \frac{2 x^{3}}{\sqrt{x^{4} + 1}} \\ ⟹ f^{'} (x) = \frac{2 x^{3}}{\sqrt{x^{4} + 1}} \end{matrix}

1c

\begin{matrix} Differentiate by definition: f (x) = \sqrt[3]{2 x + 5} \\ Solution: \\ f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = lim_{h \to 0} \frac{\sqrt[3]{2 x + 2 h + 5} - \sqrt[3]{2 x + 5}}{h} = \\ = lim_{h \to 0} \frac{2 x + 2 h + 5 - (2 x + 5)}{h ((\sqrt[3]{2 x + 2 h + 5})^{2} + \sqrt[3]{2 x + 2 h + 5} \sqrt[3]{2 x + 5} + (\sqrt[3]{2 x + 5})^{2})} = \\ lim_{h \to 0} \frac{2}{(\underset{\to \sqrt[3]{2 x + 5}}{\underset{⏟}{\sqrt[3]{2 x + 2 h + 5}}})^{2} + \underset{\to \sqrt[3]{2 x + 5}}{\underset{⏟}{\sqrt[3]{2 x + 2 h + 5}}} \sqrt[3]{2 x + 5} + (\sqrt[3]{2 x + 5})^{2}} = \frac{2}{3 (\sqrt[3]{2 x + 5})^{2}} \\ ⟹ f^{'} (x) = \frac{2}{3 (\sqrt[3]{2 x + 5})^{2}} \end{matrix}

2a

\begin{matrix} Let f be differentiable at x_{0} and f^{'} (x_{0}) = a \\ Let g be differentiable at x_{0} and g^{'} (x_{0}) = b \\ Prove by definition: f + g is differentiable at x_{0} and (f + g)^{'} (x_{0}) = a + b \\ Proof: \\ f is differentiable at x_{0} ⟹ \exists lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}} = f^{'} (x_{0}) = a \\ g is differentiable at x_{0} ⟹ \exists lim_{x \to x_{0}} \frac{g (x) - g (x_{0})}{x - x_{0}} = g^{'} (x_{0}) = b \\ ⟹ By the limit arithmetics: \\ \exists lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}} + \frac{g (x) - g (x_{0})}{x - x_{0}} = lim_{x \to x_{0}} \frac{f (x) + g (x) - (f (x_{0}) + g (x_{0}))}{x - x_{0}} = \\ = lim_{x \to x_{0}} \frac{(f + g) (x) - (f + g) (x_{0})}{x - x_{0}} = \\ \underset{By the limit arithmetics}{=} lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}} + lim_{x \to x_{0}} \frac{g (x) - g (x_{0})}{x - x_{0}} = a + b \\ ⟹ (f + g) is differentiable at x_{0} and (f + g)^{'} (x_{0}) = a + b \end{matrix}

2b

\begin{matrix} Prove by definition: \forall x \neq 0 \in R : {(\frac{1}{x})}^{'} = \frac{- 1}{x^{2}} \\ Infer by the chain rule that if g (x) is differentiable at x_{0} and g (x_{0}) \neq 0 \\ then {(\frac{1}{g})}^{'} (x_{0}) = \frac{- g^{'} (x_{0})}{(g (x_{0}))^{2}} \\ Infer by the multiplication rule that if f (x) is also differentiable at x_{0} \\ then {(\frac{f}{g})}^{'} (x_{0}) = \frac{f^{'} (x_{0}) g (x_{0}) - f (x_{0}) g^{'} (x_{0})}{(g (x_{0}))^{2}} \\ Proof: \\ {(\frac{1}{x})}^{'} = lim_{h \to 0} \frac{(\frac{1}{x + h} - \frac{1}{x})}{h} = lim_{h \to 0} \frac{\frac{x - (x + h)}{x (x + h)}}{h} = lim_{h \to 0} \frac{- 1}{x (\underset{\to x}{\underset{⏟}{x + h}})} = \frac{- 1}{x^{2}} \\ Let h (x) = \frac{1}{x} \\ Let g (x) be differentiable at x_{0} and g^{'} (x_{0}) \neq 0 \\ Then \frac{1}{g} = (h \circ g) \\ ⟹ {(\frac{1}{g})}^{'} (x_{0}) = (h \circ g)^{'} (x_{0}) = h^{'} (g (x_{0})) \cdot g^{'} (x_{0}) \underset{Given that g (x_{0}) \neq 0}{=} \frac{- 1}{(g (x_{0}))^{2}} \cdot g^{'} (x_{0}) \\ Let f (x) be differentiable at x_{0} \\ Then \frac{f}{g} = f \cdot \frac{1}{g} \\ ⟹ {(\frac{f}{g})}^{'} (x_{0}) = {(f \cdot \frac{1}{g})}^{'} (x_{0}) = f^{'} (x_{0}) \cdot (\frac{1}{g}) (x_{0}) + f (x_{0}) \cdot {(\frac{1}{g})}^{'} (x_{0}) = \\ = \frac{f^{'} (x_{0})}{g (x_{0})} + \frac{f (x_{0}) (- g^{'} (x_{0}))}{(g (x_{0}))^{2}} = \frac{f^{'} (x_{0}) g (x_{0}) + f (x_{0}) g^{'} (x_{0})}{(g (x_{0}))^{2}} \end{matrix}

2c

\begin{matrix} Prove by the inverse function theorem: (\arctan x)^{'} = \frac{1}{1 + x^{2}} \\ Proof: \\ (\arctan x)^{'} = (\tan^{- 1} x)^{'} \\ Let f (y) = \tan y \\ Then by the inverse function theorem (f^{- 1})^{'} (x) = \frac{1}{f^{'} (f^{- 1} (x))} \\ f^{'} (y) = (\tan y)^{'} = \frac{1}{\cos^{2} (y)} ⟹ f^{'} (f^{- 1} (x)) = \frac{1}{(\cos (\arctan x))^{2}} \\ \cos^{2} (y) = \frac{1}{\frac{\sin^{2} (y) + \cos^{2} (y)}{\cos^{2} (y)}} = \frac{1}{\tan^{2} (y) + 1} \\ ⟹ f^{'} (f^{- 1} (x)) = \frac{1}{\frac{1}{(\tan (\arctan x))^{2} + 1}} = x^{2} + 1 \\ ⟹ (\arctan x)^{'} = \frac{1}{f^{'} (f^{- 1} (x))} = \frac{1}{1 + x^{2}} \end{matrix}

3a

\begin{matrix} Differentiate f (x) = (((2 x + 3)^{4} + 5)^{6} + 7)^{8} + 9 \\ Solution: \\ Let h (x) = (2 x + 3)^{4} + 5 \\ Let g (x) = x^{6} + 7 \\ ⟹ f (x) = (g (h (x)))^{8} + 9 \\ h^{'} (x) = 4 (2 x + 3)^{3} \cdot (2 x + 3)^{'} = 4 \cdot (2 x + 3)^{3} \cdot 2 \\ (g (h (x)))^{'} = ((h (x))^{6} + 7)^{'} = 6 (h (x))^{5} \cdot h^{'} (x) = 6 \cdot (h (x)^{5}) \cdot 4 (2 x + 3)^{3} \cdot 2 \\ f^{'} (x) = ((g (h (x)))^{8} + 9)^{'} = 8 (g (x))^{7} \cdot g^{'} (x) = 8 \cdot (g (x))^{7} \cdot 6 \cdot (h (x)^{5}) \cdot 4 \cdot (2 x + 3)^{3} \cdot 2 \\ ⟹ f^{'} (x) = 8 \cdot (((2 x + 3)^{4} + 5)^{6} + 7)^{7} \cdot 6 \cdot ((2 x + 3)^{4} + 5)^{5} \cdot 4 \cdot (2 x + 3)^{3} \cdot 2 \end{matrix}

3b

\begin{matrix} Differentiate f (x) = \ln (\sin (\ln (\cos (x)))) \\ Solution: \\ f^{'} (x) = (\ln (\sin (\ln (\cos (x)))))^{'} = \frac{1}{\sin (\ln (\cos (x)))} \cdot (\sin (\ln (\cos (x))))^{'} = \\ = \frac{1}{\sin (\ln (\cos (x)))} \cdot \cos (\ln (\cos (x))) \cdot (\ln (\cos (x)))^{'} = \\ = \frac{1}{\sin (\ln (\cos (x)))} \cdot \cos (\ln (\cos (x))) \cdot \frac{1}{\cos (x)} \cdot (\cos (x))^{'} = - \cot (\ln (\cos (x))) \cdot \tan (x) \end{matrix}

3c

\begin{matrix} Differentiate f (x) = \frac{e^{e^{e^{x}}}}{x} \\ Solution: \\ f^{'} (x) = \frac{(e^{e^{e^{x}}})^{'} x - e^{e^{e^{x}}}}{x^{2}} \\ (e^{e^{e^{x}}})^{'} = e^{e^{e^{x}}} \cdot (e^{e^{x}})^{'} = e^{e^{e^{x}}} \cdot e^{e^{x}} \cdot (e^{x})^{'} = e^{e^{e^{x}}} \cdot e^{e^{x}} \cdot e^{x} = e^{(e^{e^{x}} + e^{x} + x)} \\ ⟹ f^{'} (x) = \frac{(e^{e^{e^{x}}} \cdot e^{e^{x}} \cdot e^{x}) x - e^{e^{e^{x}}}}{x^{2}} = \frac{e^{(e^{e^{x}} + e^{x} + x)} x - e^{e^{e^{x}}}}{x^{2}} \end{matrix}

3d

\begin{matrix} Differentiate f (x) = \sqrt[6]{7 x^{2} + 3 x - 1} \\ Solution: \\ f^{'} (x) = \frac{1}{6 (\sqrt[6]{7 x^{2} + 3 x - 1})^{5}} \cdot (7 x^{2} + 3 x - 1)^{'} = \frac{14 x + 3}{6 (\sqrt[6]{7 x^{2} + 3 x - 1})^{5}} \end{matrix}

3e

\begin{matrix} Differentiate f (x) = (\sin x)^{\cos x} \\ Solution: \\ f^{'} (x) = ((\sin x)^{\cos x})^{'} = (e^{\cos (x) \ln (\sin x)})^{'} = e^{\cos (x) \ln (\sin x)} \cdot (\cos (x) \ln (\sin x))^{'} = \\ = e^{\cos (x) \ln (\sin x)} \cdot (- \sin (x) \ln (\sin x) + \frac{\cos x}{\sin x} \cdot (\sin x)^{'}) = \\ = (\sin x)^{\cos x} \cdot (\cot (x) \cos (x) - \sin (x) \ln (\sin x)) \end{matrix}

3f

\begin{matrix} Differentiate f (x) = \sin (x^{\cos x}) \\ Solution: \\ f^{'} (x) = \cos (x^{\cos x}) \cdot (x^{\cos x})^{'} \\ (x^{\cos x})^{'} = (e^{\cos (x) \ln (x)})^{'} = e^{\cos (x) \ln (x)} \cdot (\cos (x) \ln (x))^{'} = x^{\cos x} \cdot (- \sin (x) \ln (x) + \frac{\cos x}{x}) \\ ⟹ f^{'} (x) = \cos (x^{\cos x}) \cdot x^{\cos x} \cdot (\frac{\cos x}{x} - \sin (x) \ln (x)) \end{matrix}

3g

\begin{matrix} Find f^{(3)} (x) where f (x) = e^{e^{x}} \\ Solution: \\ f^{'} (x) = (e^{e^{x}})^{'} = e^{e^{x}} \cdot (e^{x})^{'} = f (x) \cdot e^{x} \\ ⟹ f^{″} (x) = (f^{'} (x))^{'} = (f (x) \cdot e^{x})^{'} = f^{'} (x) \cdot e^{x} + f (x) \cdot (e^{x})^{'} = f (x) \cdot e^{x} \cdot e^{x} + f (x) \cdot e^{x} \\ ⟹ f^{″} (x) = f^{'} (x) \cdot e^{x} + f^{'} (x) \\ ⟹ f^{(3)} (x) = (f^{″} (x))^{'} = (f^{'} (x) \cdot e^{x} + f^{'} (x))^{'} = (f^{'} (x) \cdot e^{x})^{'} + f^{″} (x) = \\ f^{″} (x) \cdot e^{x} + f^{'} (x) \cdot (e^{x})^{'} + f^{″} (x) = f^{″} (x) \cdot e^{x} + f^{″} (x) + f^{'} (x) \cdot e^{x} \\ ⟹ f^{(3)} (x) = (f (x) \cdot e^{x} \cdot e^{x} + f (x) \cdot e^{x}) \cdot e^{x} + f (x) \cdot e^{x} \cdot e^{x} + f (x) \cdot e^{x} + f (x) \cdot e^{x} \cdot e^{x} = \\ = f (x) \cdot e^{3 x} + f (x) \cdot e^{2 x} + f (x) \cdot e^{2 x} + f (x) \cdot e^{2 x} + f (x) \cdot e^{x} = \\ = e^{e^{x}} \cdot e^{3 x} + 3 e^{e^{x}} \cdot e^{2 x} + e^{e^{x}} \cdot e^{x} \end{matrix}

3h

\begin{matrix} Find f^{(2022)} (x) where f (x) = \sin (2 x + 3) + 2 e^{x} + 7 x \\ Solution: \\ f^{(2022)} (x) = (\sin (2 x + 3) + 2 e^{x} + 7 x)^{(2022)} = \\ = (\sin (2 x + 3))^{(2022)} + (2 e^{x})^{(2022)} + (7 x)^{(2022)} \\ (7 x)^{(2022)} = ((7 x)^{″})^{(2020)} = (0)^{(2020)} = 0 \\ (2 e^{x})^{(2022)} = 2 (e^{x})^{(2022)} = 2 e^{x} \\ (\sin (2 x + 3))^{'} = 2 \cos (2 x + 3) \\ ⟹ (\sin (2 x + 3))^{″} = - 2^{2} \sin (2 x + 3) \\ ⟹ (\sin (2 x + 3))^{(3)} = - 2^{3} \cos (2 x + 3) \\ ⟹ (\sin (2 x + 3))^{(4)} = 2^{4} \sin (2 x + 3) \\ ⟹ (\sin (2 x + 3))^{(4 k)} = 2^{4 k} \sin (2 x + 3) \\ 2022 = 4 \cdot 505 + 2 \\ ⟹ (\sin (2 x + 3))^{(2022)} = ((\sin (2 x + 3))^{(4 \cdot 505)})^{″} = (2^{2020} \sin (2 x + 3))^{″} = \\ = - 2^{2022} \sin (2 x + 3) \\ ⟹ f^{(2022)} (x) = - 2^{2022} \sin (2 x + 3) + 2 e^{x} \end{matrix}

4a

\begin{matrix} Let f (x) = {\begin{cases} \sin^{2} (x) \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases} \\ Is f continuous on R ? \\ Is f differentiable on R ? \\ Is f^{'} continuous on R ? \end{matrix}

\begin{matrix} Solution: \\ \sin^{2} (x) is continuous on R \\ \sin (\frac{1}{x}) is continuous on R ∖ {0} \\ ⟹ \sin^{2} (x) \sin (\frac{1}{x}) is continuous on R ∖ {0} \\ ⟹ f is continuous on R ∖ {0} \\ lim_{x \to 0} f (x) = lim_{x \to 0} \underset{\to 0^{2}}{\underset{⏟}{\sin^{2} (x)}} \cdot \underset{- 1 \leq \sin (\frac{1}{x}) \leq 1}{\underset{⏟}{\sin (\frac{1}{x})}} = 0 = f (0) \\ ⟹ f is continuous at 0 ⟹ f is continuous on R \\ lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = \frac{\sin^{2} (x) \sin (\frac{1}{x})}{x} = lim_{x \to 0} \frac{\sin x}{x} \cdot lim_{x \to 0} \sin (x) \sin (\frac{1}{x}) = 1 \cdot 0 = 0 \\ ⟹ f is differentiable at 0 and f^{'} (0) = 0 \\ Let x \neq 0 \\ ⟹ f (x) = \sin^{2} (x) \sin (\frac{1}{x}) which is differentiable at x \\ and f^{'} (x) = (\sin^{2} (x))^{'} \sin (\frac{1}{x}) + \sin^{2} (x) {(\sin (\frac{1}{x}))}^{'} \\ ⟹ f^{'} (x) = 2 \sin (x) \cos (x) \sin (\frac{1}{x}) - \frac{\sin^{2} (x) \cos (\frac{1}{x})}{x^{2}} \\ ⟹ f^{'} (x) = {\begin{cases} 2 \sin (x) \cos (x) \sin (\frac{1}{x}) - \frac{\sin^{2} (x) \cos (\frac{1}{x})}{x^{2}} & x \neq 0 \\ 0 & x = 0 \end{cases} \\ 2 \sin (x) \cos (x) \sin (\frac{1}{x}) - \frac{\sin^{2} (x) \cos (\frac{1}{x})}{x^{2}} is continuous on R ∖ {0} \\ ⟹ f^{'} (x) is continuous on R ∖ {0} \\ lim_{x \to 0} f (x) = lim_{x \to 0} 2 \underset{\to 0}{\underset{⏟}{\sin (x)}} \underset{\to 1}{\underset{⏟}{\cos (x)}} \underset{- 1 \leq \sin (\frac{1}{x}) \leq 1}{\underset{⏟}{\sin (\frac{1}{x})}} - \frac{\sin^{2} (x) \cos (\frac{1}{x})}{x^{2}} = \\ = - lim_{x \to 0} \underset{\to 1^{2} = 1}{\underset{⏟}{{(\frac{\sin x}{x})}^{2}}} \cos (\frac{1}{x}) = - lim_{x \to 0} \cos (\frac{1}{x}) which does not exist \\ ⟹ f^{'} is not continuous at 0 ⟹ f^{'} is not continuous on R \end{matrix}

4b

\begin{matrix} Let f be a continuous on R \\ Let g (x) = {\begin{cases} \frac{f (x) \sin^{2} (x)}{x} & x \neq 0 \\ 0 & x = 0 \end{cases} \\ Is g continuous at 0 ? Is g differentiable at 0 ? \\ Solution: \\ lim_{x \to 0} g (x) = lim_{x \to 0} \frac{f (x) \sin^{2} (x)}{x} = lim_{x \to 0} \underset{\to 1}{\underset{⏟}{\frac{\sin x}{x}}} \cdot lim_{x \to 0} f (x) \sin (x) = lim_{x \to 0} f (x) \sin (x) \\ f is continuous at 0 ⟹ lim_{x \to 0} f (x) = f (0) \in R \\ ⟹ lim_{x \to 0} f (x) \sin (x) = lim_{x \to 0} f (x) \cdot lim_{x \to 0} \sin (x) = f (0) \cdot 0 = 0 ⟹ lim_{x \to 0} g (x) = 0 = g (0) \\ ⟹ g is continuous at 0 \\ lim_{x \to 0} \frac{g (x) - g (0)}{x - 0} = lim_{x \to 0} \frac{f (x) \sin^{2} (x)}{x^{2}} = lim_{x \to 0} f (x) \cdot lim_{x \to 0} {(\frac{\sin x}{x})}^{2} = f (0) \cdot 1^{2} = f (0) \\ ⟹ g is differentiable at 0 \end{matrix}

5a

\begin{matrix} Calculate \sqrt[3]{26} \\ Solution: \\ Let f (x) = \sqrt[3]{x} \\ f^{'} (x) = \frac{1}{3 (\sqrt[3]{x})^{2}} \\ f (27) = 3 \\ f^{'} (27) = \frac{1}{3 \cdot 3^{2}} = \frac{1}{27} \\ ⟹ The "slope" equation at (27, 3) is y - 3 = \frac{1}{27} (x - 27) \\ ⟹ y = \frac{x}{27} + 2 \\ Let x = 26 \\ ⟹ \sqrt[3]{26} \approx y = 2 + \frac{26}{27} \\ Let g (x) = \ln (x) \\ g^{'} (x) = \frac{1}{x} \\ g (1) = 0 \\ g^{'} (1) = \frac{1}{1} = 1 \\ ⟹ The "slope" equation at (1, 0) is y - 0 = 1 (x - 1) \\ ⟹ y = x - 1 \\ Let x = 1.002 \\ ⟹ \ln (1.002) \approx y = 1.002 - 1 = 0.002 \end{matrix}

5b

\begin{matrix} Prove: \forall x > 0 : \ln (1 + x) < x \\ Proof: \\ Let x > 0 \\ Let f (x) = x - \ln (1 + x) \\ f (0) = 0 \\ f^{'} (x) = 1 - \frac{1}{1 + x} \\ \frac{1}{1 + x} is monotonically decreasing on [0, \infty) \\ ⟹ f^{'} (x) is monotonically increasing on [0, \infty) \\ ⟹ \forall x > 0 : f (x) > f (0) ⟹ x - \ln (1 + x) > 0 ⟹ x > \ln (1 + x) \end{matrix}

5c

\begin{matrix} Find values of a such that f (x) = a x^{3} + 4 a x^{2} + 8 x has two local extremums \\ Solution: \\ f has a local extremum at x_{0} ⟹ f^{'} (x_{0}) = 0 \\ f has two local extremums ⟹ f^{'} (x) = 0 has two solutions \\ f^{'} (x) = 3 a x^{2} + 8 a x + 8 \\ f^{'} (x) = 0 ⟺ 3 a x^{2} + 8 a x + 8 = 0 \\ This equation has two solutions when: \\ D = 64 a^{2} - 96 a > 0 \\ 64 a^{2} - 96 a = 32 a (2 a - 3) \\ a (2 a - 3) > 0 ⟺ [\begin{matrix} a < 0 \\ a > \frac{3}{2} \end{matrix} \\ Let us now check if these two critical points are in fact extremums \\ f^{″} (x_{0}) > 0 ⟹ x_{0} is a local minimum \\ f^{″} (x_{0}) < 0 ⟹ x_{0} is a local maximum \\ f^{″} (x) = 6 a x + 8 a = 2 a (3 x + 4) \\ f^{″} (x) = 0 ⟺ x = - \frac{4}{3} \\ f^{'} (- \frac{4}{3}) = 3 a (\frac{16}{9}) - 8 a (\frac{4}{3}) + 8 = \frac{16 a}{3} - \frac{32 a}{3} + 8 = 8 - \frac{16 a}{3} \\ f^{'} (- \frac{4}{3}) = 0 ⟺ a = \frac{3}{2} ⟺ f^{'} (x) = 0 has only one solution \\ ⟹ \forall a < 0 or a > \frac{3}{2} : f^{″} (x) \neq 0 \\ ⟹ [\begin{matrix} a < 0 \\ a > \frac{3}{2} \end{matrix} ⟹ f has two local extremums \end{matrix}

6a

\begin{matrix} Find local extremums of f (x) = \sqrt[x]{x} for x > 0 \\ Solution: \\ f (x) = x^{1 / x} = e^{\ln (x) / x} \\ ⟹ f^{'} (x) = e^{\ln (x) / x} \cdot {(\frac{\ln (x)}{x})}^{'} = \sqrt[x]{x} \cdot \frac{\frac{1}{x} x - \ln (x)}{x^{2}} = \sqrt[x]{x} \frac{1 - \ln (x)}{x^{2}} \\ f^{'} (x) = 0 ⟺ \ln (x) = 0 ⟺ x = e \\ x > e ⟹ 1 - \ln (x) < 0 ⟹ f^{'} (x) < 0 ⟹ f (x) < f (e) \\ x < e ⟹ 1 - \ln (x) > 0 ⟹ f^{'} (x) > 0 ⟹ f (x) < f (e) \\ ⟹ x = e is a local maximum of f \end{matrix}

6b

\begin{matrix} Explain why the local maximum from 6a is also a global maximum of f \\ Explanation: \\ Let x < e \\ x < e ⟹ 1 - \ln (x) > 0 ⟹ f^{'} (x) > 0 ⟹ f (x) < f (e) \\ Let x > e \\ x > e ⟹ 1 - \ln (x) < 0 ⟹ f^{'} (x) < 0 ⟹ f (x) < f (e) \\ ⟹ \forall x \in R ∖ {e} : f (x) < f (e) \\ ⟹ x = e is also a global maximum of f \end{matrix}

6c

\begin{matrix} Determine which number is larger: π^{e} or e^{π} \\ Solution: \\ By 6b \forall x > 0 : f (x) < f (e) ⟹ x^{1 / x} < e^{1 / e} \\ Let x = π \\ π^{1 / π} < e^{1 / e} ⟹ π < e^{π / e} ⟹ π^{e} < e^{π} \end{matrix}