Calculus - Typical problems

A1

1. Draw the graph first!
   1. As we will need the derivatives of $f(x)=\frac{1}{x^2+4}=(x^2+4)^{-1}$, let's calculate them here: $$\begin{array}{lll} f'(x)&=&-(x^2+4)^{-2}\cdot 2x\\ &=&-\frac{2x}{(x^2+4)^2}\\ \end{array}$$ To find the second derivative, we apply the product rule to $f'(x)=\underbrace{-(x^2+4)^{-2}}_{=u(x)}\cdot \underbrace{2x}_{=v(x)}$: $$\begin{array}{lll} f''(x)&=& \underbrace{2(x^2+4)^{-3}\cdot 2x}_{=u'(x)}\cdot \underbrace{2x}_{=v(x)} -\underbrace{(x^2+4)^{-2}}_{=u(x)}\cdot \underbrace{2}_{=v'(x)}\\ &=& 8x^2 (x^2+4)^{-3}-2(x^2+4)^{-2}\\ \end{array}$$ Note that to determine the derivative of $u(x)=-(x^2+4)^{-2}$ we have to apply the chain rule, where the inner function is $i(x)=x^2+4$ and the outer function is $o(x)=x^{-2}$.
      1. $x$-Intercept: There is no $x$ with $f(x)=0\rightarrow \frac{1}{x^2+4}=0$, as $x^2+4>0$ for all $x$. So no $x$-intercept.
      2. $y$-intercept: $f(0)=0.25$
      3. stationary points: Find $x$ with $f'(x)=0$, that is $$-(x^2+4)^{-2}\cdot 2x=0$$ This is only possible for $x=0$. So $P(0|0.25)$ is the only stationary point. Because of $f''(0)=-0.125<0$ we see that $P$ is a local maximum.
      4. Inflection point: Find $x$ with $f''(x)=0$, thus $$\begin{array}{lll} 8x^2 (x^2+4)^{-3}-2(x^2+4)^{-2}&=&0 \quad|\cdot (x^2+4)^3\\ 8x^2-2(x^2+4)&=&0\\ 6x^2-8&=&0\\ x_{1,2}&=&\pm\sqrt{\frac{4}{3}} \end{array}$$ We actually would have to check if $f'''(x_{1,2})\neq 0$, but we will skip this step. Actually it is the case. So $Q_{1,2} (\pm\sqrt{\frac{4}{3}}|\frac{3}{16})$ are the inflection points.
   2. As the graph is always above the $x$-axis, we can simply take the integral for finding the area. From the hint we know that $g$ is an antiderivative of $f$ (see 2). Thus we have $$A=\int_{-\sqrt{4/3}}^{\sqrt{4/3}} f(x)\, dx = g(\sqrt{4/3})-g(-\sqrt{4/3})=\frac{\pi}{6}$$
2. It is
   1. $x$-intercept: find $x$ with $x^3e^{-2x}=0$ As $e^{-2x}>0$ for all $x$, it must be $x^2=0 \rightarrow x=0$. So there is one $x$-intercept at $x=0$.
   2. Stationary points: find $x$ with $f'(x)=0$: $$f'(x)=3x^2e^{-2x}+x^3e^{-2x}(-2)=e^{-2x}(3x^2-2x^3)=0$$ As $e^{-2x}>0$ for all $x$, it is $3x^2-2x^3=0$. Factoring out $x^2$, we get $$x^2(3-2x)=0$$ and it follows $x_1=0$ and $x_2=1.5$. The stationary points are therefore $P_1(0|f(0))=P_1(0|0)$ and $P_2(1.5|f(1.5))=P_2(1.5|0.168)$
3. It is
   1. $x$-intercept: $\sin(2x)+\cos(2x)=0 \rightarrow \sin(2x)=-\cos(2x)$. Divide both sides of the equation by $\cos(2x)$ $$\frac{\sin(2x)}{\cos(2x)}=-1$$ $$\tan(2x)=-1$$ Thus $$2x=\arctan(-1)=-\frac{\pi}{4} \rightarrow x=-\frac{\pi}{8}$$
   2. Stationary point: $f^\prime(x)=2\cos(2x)-2\sin(2x)=0$, also $$\tan(2x)=1 \rightarrow x=\frac{\pi}{8} \rightarrow y=f(\frac{\pi}{8})=\sqrt{2}$$
   3. Inflection point: $f^{\prime\prime}(x)=-4\sin(2x)-4\cos(2x)=0$, also $$\tan(2x)=-1 \rightarrow x=-\frac{\pi}{8}$$

A2

• Step 1: Let's find the point of intersection, that is, find $x$ with $f_c(x)=g_c(x)$, that is, $$\begin{array}{lll} x^2+xc&=&x^2+c\\ xc&=&c\\ x&=&1 \end{array}$$ It follows $x=1$ (for $c\neq 0$). So independent of the value of $c$, the two graphs always intersect at $x=1$.
• Step 2: The slope of $f_c$ at $x$ is $$f_c'(x)=2x+c$$ and the slope of $g_c$ at $x$ is $$g'_c(x)=2x$$ (you must treat $c$ as a normal number). Thus, at $x=1$, the slope of the tangents are $$f_c'(1)=2+c \text{ and } g'_c(1)=2$$ To find $c$ such that the two tangents intersect at a right angle, we need to find $c$ such that (see hint) $$2+c=-\frac{1}{2}$$ and it follows $c=-2.5$.

A3

$f(x)=-\frac{1}{36\sqrt{3}}x^2(x-6)$

A4

Draw the graph of $f(x)=\sqrt{x}$, chose an $x$ and draw the point $P$ on the graph of $f$ at $x$. It has the coordinates $P(x|\sqrt{x})$.

From the theorem of Pythagoras follows that

and thus we have

So find $x$ which minimises $d$, that is, find a local minimum of $d$, that is, find $x$ with $f'(x)=0$ (the gives us the stationary points).

Applying the chain rule to find the derivative of $f$, we get

And because the first factor can never be zero, the only possible solution is that

Plot the graph of $d$ to determine if there is a local minimum at $x=3.5$, or (more difficult), use the second derivative of $d$ to show that $f''(3.5)>0$. It actually is a local minimum, so we get $\underline{P(3.5|\sqrt{3.5})}$

A5

$V=\pi \int_1^5 \left(\frac{1}{\sqrt{x}}\right)^2\, dx = \pi\cdot (\ln(5)-\ln(1))=1.6\pi$

A6

$f_{upper}(x)=a(x-3)^2$ and because of $f_{upper}(0)=-3$ we get $a=-\frac{1}{3}$, so we have

$f_{lower}(x)=a(x-3)(x+3)$ and because of $f_{lower}(0)=-3$ we get $a=\frac{1}{3}$, so we have

To find the area, we need to integrate, so let's find the antiderivatives of $f_{upper}$ and $f_{lower}$ first:

and

The area $A$ of the shaded region is then

A7

We can calculate the two solids of revolution of $f$ and $g$ and subtract one from the other. First, we need the intersection points:

Which has the solutions $x_1 = 0$ and $x_2 = 2$. If we call the volumes of the solids of revolution $V_f$ and $V_g$, respectively, we get:

The resulting volume is $V_f- V_g = \frac{2^8}{35}\pi$.

A8

The intersection point is at $S(\frac{1}{3},\frac{4}{9})$, the angle of intersection (angle between tangents at $f$ and $g$ at $S$) is $\alpha=122.57^\circ$ (or $180^\circ-\alpha=57.43^\circ$).