Calculus - Basic functions 2

Solution 1

We denote the antiderivative by $F$, the $y$-intercept by $y_0$ and the $x$-intercept(s) by $x_0, x_1, ...$

1a)

$f(x)=2x+1$, $y_0=1, x_0=-0.5$, $f^\prime(x)=2$, $F(x)=x^2+x$

1b)

$f(x)=x^2$, $x_0=y_0=0$, $f'(x)=2x$, $F(x)=\frac{1}{3}x^3$

1c)

$f(x)=e^x$, $y_0=1$, $f'(x)=e^x$, $F(x)=e^x$

1d)

$f(x)=x^{-2}=\frac{1}{x^2}$, $f'(x)=-2x^{-3}=-\frac{2}{x^3}$, $F(x)=-x^{-1}=-\frac{1}{x}$

1e)

$f(x)=x^{-3}=\frac{1}{x^3}$, $f'(x)=-3x^{-4}=-\frac{3}{x^4}$, $F(x)=-\frac{1}{2}x^{-2}=-\frac{1}{2x^2}$

1f)

$f(x)=\ln(x)$, $x_0=1$, $f'(x)=\frac{1}{x}$, $F(x)\text{ skip, not required}$

1g)

$f(x)=2=2x^0$, $y_0=2$, $f'(x)=0$, $F(x)=2x$

1h)

$f(x)=\sin(x)$, $y_0=0$, $x_0=0, x_1=\pm\pi, x_2=\pm 2\pi, ...$, $f'(x)=\cos(x)$, $F(x)=-\cos(x)$

1i)

$f(x)=x^{1/2}=\sqrt{x}$, $x_0=y_0=0$, $f'(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$, $F(x)=\frac{2}{3}x^{3/2}=\frac{2}{3}\sqrt{x^3}$

1j)

$f(x)=x^{1/3}=\sqrt[3]{x}$, $x_0=y_0=0$, $f'(x)=\frac{1}{3}x^{-2/3}=\frac{1}{3\sqrt[3]{x^2}}$, $F(x)=\frac{3}{4}x^{4/3}=\frac{3}{4}\sqrt[3]{x^4}$

1k)

$f(x)=\tan(x)$, $y_0=0, x_0=0, x_1=\pm\pi, x_2=\pm 2\pi, ...$ (because $\tan(x)=\frac{\sin(x)}{\cos(x)}$).

To find the derivative, we apply the product rule and chain rule to the function

thus

Calculating the antiderivative is more difficult, and you do not have to know how to do this.

1l)

$f(x)=x^3-4x$, $y_0=f(0)=0$. To find $x_0$, we write $f(x)=x(x^2-4)=0$, and it follows $x_0=0, x_1=\pm 2$. $f'(x)=3x^2-4$, $F(x)=\frac{1}{4}x^4-4\frac{1}{2}x^2=\frac{1}{4}x^4-2x^2$

Solution 2

2f)

Polynomial: $f(x)=a(x+1)(x-1)(x-3)$. From $f(0)=3$ follows $a(1)(-1)(-3)=3\rightarrow a=1$.

2g)

Polynomial or quadratic function (the graph is a parabola): $g(x)=a(x+4)^2$ (the square because the graph touches the $x$-axis at $-4$). Because of $g(-2)=-2$ follows $a(2)^2=-2\rightarrow a=-0.5$

2h)

Polynomial: $h(x)=a(x-4)^2(x-6)^2$ (squares because of touch points). Because $h(5)=2$ follows $a(-1)^2(1)^2)=2\rightarrow a=2$

2i)

Sine function, but scaled in $x$-direction by some factor $u$, $f(x)=2\sin(ux)$:

Because we know that

choose $u$ such that

thus it is $u=\frac{\pi}{2}$. It is $f(x)=\sin(\frac{\pi}{2}x)$. Note that the graph is also stretched in $y$-direction by factor $2$, so we have the final result

2k)

Exponential function passing through the points $A(0|2)$ and $B(2|3)$. If we interpret these points as an exponential growth process,

We get a growth rate of $u=3/2=1.5$, and thus we have

2m)

Linear function: $m(x)=ax+b$ with slope $a=\frac{-1}{2}=-0.5$. Thus we have

To find $b$, we note that $m(1)=5$, and thus $-0.5\cdot 1+b=5$, and thus $b=5.5$. It is

2l)

Quadratic function.

Approach 1: Because of lack of $x$-intercepts, we take the approach

We have (see graph)

From equation (1) follows $c=2-9a+3b$ and inserting this into equations (2) and (3) we get

And from these two equations follows $b=2+5a$ and therefore

and thus $b=2+5\cdot 2=12$ and $c=2-9\cdot 2+3\cdot 12=20$. Thus we have

Approach 2: We shift the graph down by $2$, so that it touches the $x$-axis at $-3$. This graph also passes through the point $(-2|2)$. Thus, for this shifted graph we have

and

Thus we have $L(x)=2(x+3)^2$. Shifting the graph up by $2$, we then get

Indeed, if we expand, we get the result from the first approach:

Solution 3

1. stretch of graph $f$ in $y$-direction by factor $2$
2. shift up of graph $f$ by $1$
3. reflect graph $f$ about $x$-axis
4. shift right of graph $f$ by $1$

Solution 4

1. All $x$ on one side, numbers on other side: $2x=3 \rightarrow x=1.5$
2. Quadratic equation, so midnight formula, or simpler, factor out $x$: $x(2x-3)=0 \rightarrow x_1=0, x_2=1.5$
3. Quadratic equation, midnight formula: $2x^2-6x+4=0 \rightarrow x_{1,2}=\frac{6\pm\sqrt{36-32}}{4}\rightarrow x_1=2, x_2=1$
4. Factor out $2x$: $2x(x^2-2)=0 \rightarrow x_1=0, x_2=\pm \sqrt{2}$
5. Divide both sides by $0.5$: $\sqrt{x}=x-2$ and then square both sides: $x=(x-2)^2=x^2-4x+4$. This is a quadratic equation: $x^2-5x+4=0$. Using the midnight formula, we get $x_1=1$ and $x_2=4$. Inserting into the original equation, we see that only $x=4$ works.
6. Subtract both sides by $1$, and then divide both sided by $2$: $x^{2/3}=9$. Then raise both sides by $\frac{3}{2}$: $$\underbrace{(x^{2/3})^{3/2}}_{x}=\underbrace{9^{3/2}}_{27}$$ It follows $x=27$ (another solution is $x=-27$).
7. Divide both sides by $3$: $2^{(x-2)/3}=8$. Apply $\log_2(.)$ on both sides: $$\underbrace{\log_2(2^{(x-2)/3})}_{(x-2)/3\cdot\log_2(2)} =\underbrace{\log_2(8)}_{3}$$ Because of $\log_2(2)=1$ and $\log_2(8)=3$ we get the equation $$\frac{x-2}{3}=3$$ and it follows $x=11$.
8. Divide both sides by 4: $\log_{10}(2x+1)=2 \rightarrow 10^2=2x+1$, thus it follows $x=49.5$
9. As $e^{(...)}>0$, it must by $x^2-1=0\rightarrow x=\pm 1$.