Week 3 - ex 4 a.o.

1 files changed, 71 insertions, 5 deletions

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     \item \begin{enumerate}
             \item $f(x) = \log_32 + \log_3x$. Then $f'(x) = \frac1{x\ln3}$. Then $f''(x) = -\frac{\ln3}{x^2(\ln3)^2} = -\frac1{x^2\ln3}$. Finally $f'''(x) = -\frac{-2\ln3\cdot x}{x^4\cdot(\ln3)^2} = \frac2{x^3\ln3}$.
-            \item %todo
+
+            \item Let's look at $f(x) = k\cdot\cos(c\cdot x)$. We can find the following derivatives:
+                \begin{align*}
+                    f^{(1)}(x)  &= k\cdot-\sin(cx)\cdot c = -ck\cdot\sin(cx).\\
+                    f^{(2)}(x)  &= -ck\cdot\cos(xc)\cdot c = -c^2k\cdot\cos(cx).\\
+                    f^{(3)}(x)  &= -c^2k\cdot-\sin(cx)\cdot c = c^3k\cdot\sin(cx).\\
+                    f^{(4)}(x)  &= c^3k\cdot\cos(cx)\cdot c = c^4k\cdot\cos(cx).\\
+                \end{align*}
+                Note that $f^{(4)}$ is also of the form $k\cdot\cos(c\cdot x)$. We can thus write for any $f$ of that form: $f^{(4n)}(x) = c^{4n}k\cdot\cos(cx)$ with $n\in\mathbb N$.
+
+                If we fill in $f(x) = g(x)$ and $n=\frac{2012}{4}$, we find $g^{(2012)}(x) = 3^{2012}\cos(3x)$. Then from this we find:
+                \begin{align*}
+                    g^{(2013)}(x)   &= -3^{2013}\sin(3x).\\
+                    g^{(2014)}(x)   &= -3^{2014}\cos(3x).\\
+                    g^{(2015)}(x)   &= 3^{2015}\sin(3x).
+                \end{align*}
         \end{enumerate}
 
     \item \begin{enumerate}
             \item $x-3$, $x+1$ and $x^2$ are all from $\mathbb R$ to $\mathbb R$. Multiplication is from $\mathbb R\times\mathbb R$ to $\mathbb R$. Therefore, $f : \mathbb R \to \mathbb R$. 
+
             \item To find the roots we equate $f(x)$ with zero and solve for $x$. That gives $x=-1$ or $x=3$, so we have $(-1,0)$ and $(3,0)$ as roots of $f$.
 
                 To find the y-intercept, we fill in $x=0$ in $f(x)$: $(0+1)^2(0-3)=-3$. This gives $(0,-3)$.
-            \item %todo
+
+            \item $\lim_{x\to\infty}(x+1)^2(x-3)=\infty$, since the limits of $x\to\infty$ for $(x+1)^2$ and $x-3$ are both $\infty$.
+
+                $\lim_{x\to-\infty}(x+1)^2(x-3)=\lim_{x\to\infty}x^3-x^2-5x-3$. $x^3$ is dominant (because it has the highest exponent), so this limit is $\lim_{x\to-\infty}x^3=-\infty$.
+
             \item $f'(x) = \left((x+1)^2(x-3)\right)' = \left(x^3-x^2-5x-3\right)' = 3x^2 - 2x - 5$.
 
                 $f''(x) = \left(3x^2 - 2x - 5\right)' = 6x - 2$.
+
+            \item $f'(x) = 3x^2-2x-5 = 0$ gives $x=\frac{2 + \sqrt{4+60}}{6} = \frac53$ or $x=\frac{2 - \sqrt{4+60}}{6} = -1$. This gives the points $(\frac53,0)$ and $(-1,0)$.
+
+                $f''(x) = 6x-2 = 0$ gives $x=\frac13$. This gives $(\frac13,0)$.
+
+            \item We already found the $x$s, namely those $x$ where $f'(x)=0$, i.e. $x=-1$ and $x=\frac53$. We then get the critical points $(-1,f(-1)) = (-1,0)$ and $(\frac53,f(\frac53)) = (\frac53, -4\cdot(\frac43)^3)$.
+
+            \item Again we use those $x$ where $f'(x)=0$.
+
+                For $x=\frac53$ we have $f''(x) = 8 > 0$, therefore this (the critical point with this $x$) is a local minimum.
+
+                For $x=-1$ we have $f''(x) = -8 < 0$, therefore this is a local maximum.
+
+            \item $f''(x) > 0$ leads to $x > \frac13$, so $f$ is convex on $(\frac13,\infty)$. $f''(x) < 0$ leads to $x < \frac13$, so $f$ is concave on $(-\infty,\frac13)$. A point of inflection (the only) would then be $(\frac13, f(\frac13)) = (\frac13, \frac{128}{27})$.
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item The function in the numerator goes from $\mathbb R$ to $\mathbb R$, as does the function in the denominator. However, the fraction requires this denominator to be unequal to zero. We thus find as domain $\{x\in\mathbb R \mid x+2\neq 0\} = \mathbb R\setminus\{-2\}$.
+                
+            \item For the roots of $f$ we solve $f(x)=0$ for $x$. That gives $(x-2)^2=0$, so $(x-2)=0$ and thus $x=2$. This yields the point $(2,0)$.
+
+                The graph of $f$ intersect the $y$ axis at $x=0$; this gives the point $(0,f(0)) = (0, 2)$.
+                
             \item %todo
+                
+            \item \begin{align*}
+                    f(x)    &= \frac{(x-2)^2}{x+2} = \frac{x^2 - 4x + 4}{x+2}.\\
+                    f'(x)   &= \frac{(x+2)(2x-4)-(x^2-4x+4)(1)}{(x+2)^2}\\
+                            &= \frac{x^2 + 4x - 12}{(x+2)^2}\\
+                            &= \frac{(x-2)(x+6)}{(x+2)^2}\\
+                            &= \frac{x^2+4x-12}{x^2+4x+4}.\\
+                    f''(x)  &= \frac{(x^2+4x+4)(2x+4)-(x^2+4x-12)(2x+4)}{(x+2)^4}\\
+                            &= \frac{32}{(x+2)^3}.
+                \end{align*}
+                
+            \item Again, we solve $f'(x)=0$ and $f''(x)=0$ for $x$:
+                \begin{align*}
+                    f'(x)   &= 0\\
+                    \frac{(x-2)(x+6)}{(x+2)^2} &= 0
+                \end{align*}
+                This gives $x=2$ or $x=-6$, with the points $(2,0)$ and $(-6,0)$.
+
+                It is obvious at first sight that $f''(x)$ has no zeroes: it numerator can never be equal to zero.
+                
+            \item We take $\{(x,f(x)) \mid f'(x)=0\} = \{(2,f(2)),(-6,f(-6))\} = \{(2,0),(-6,-16)\}$.
+                
+            \item At $x=2$ we have $f''(x)=\frac12>0$, so $(2,0)$ is a local minimum. At $x=-6$ we have $f''(x)=-\frac12<0$, so $(-6,-16)$ is a local maximum.
+                
             \item %todo
+                
             \item %todo
-            \item %todo
+                
         \end{enumerate}
 
     \item %todo
 
-    \item %todo
-
 \end{enumerate}
 
 \end{document}