heureka

Consider the function g(x)=3x-4. 

For what value of  a  is g(a)=0 ?

\(g(a) = 0:\)

\(\begin{array}{|rcll|} \hline g(a)= 0 &=&3a-4 \\ 0&=& 3a-4 \quad & | \quad +4 \\ 4&=& 3a \quad & | \quad :3 \\ \dfrac{4}{3} &=& a \\ \mathbf{ a } &\mathbf{=}& \mathbf{ \dfrac{4}{3} } \\ \hline \end{array} \)

If j, k, and l are positive with jk = 24, jl = 48, and kl = 18,

find j + k + l.

1.

\(\begin{array}{|rcll|} \hline jk\cdot jl \cdot kl = (jkl)^2 &=& 24\cdot 48\cdot 18 \\ &=& 20736 \\ jkl &=& \sqrt{20736} \\ \mathbf{jkl} &\mathbf{=}& \mathbf{ 144} \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline j+k+l &=& \dfrac{jk\cdot jl}{jkl} + \dfrac{jk\cdot kl}{jkl} + \dfrac{jl\cdot kl}{jkl} \\ &=& \dfrac{jk\cdot jl+jk\cdot kl+jl\cdot kl}{jkl} \\ &=& \dfrac{24\cdot 48+24\cdot 18+48\cdot 18}{144} \\ \mathbf{j+k+l} &\mathbf{=}& \mathbf{ 17} \\ \hline \end{array}\)

Positive real numbers r, s satisfy the equations

r^2+s^2 = 1 and

r^4 + s^4 = 7/8.

Find rs.

\(\begin{array}{|rcll|} \hline (r^2+s^2)^2 &=& r^4+2r^2s^2+s^4 \\ (r^2+s^2)^2 &=& r^4+s^4+2r^2s^2 \\ 2r^2s^2 &=& (r^2+s^2)^2-(r^4+s^4) \quad & | \quad r^2+s^2 = 1 \qquad r^4 + s^4 = \dfrac{7}{8} \\ 2r^2s^2 &=& 1^2-\dfrac{7}{8} \\ 2r^2s^2 &=& \dfrac{1}{8} \\ r^2s^2 &=& \dfrac{1}{16} \\ \mathbf{rs} &\mathbf{=}& \mathbf{\dfrac{1}{4}} \\ \hline \end{array}\)

Find the fourth arithmetic means between -21 and -36

\(\text{We use $a_n = a_1 + (n-1)d$ to find the common difference $d$.} \)

Solution

\(\text{The first term is $a_1 = - 21$ and the fourth term is $a_4 = - 36 $.}\\ \text{We must find the common difference so that the terms }\)

\(\begin{array}{cccc} -21, & -21+d, & -21+2d, & -36 \\ \uparrow & \uparrow & \uparrow & \uparrow \\ a_1 & a_2 & a_3 & a_4 \end{array}\)

\(\text{form an arithmetic sequence.} \)

\(\text{To find the common difference $d$, we substitue $-21$ for $a_1$; 4 for $n$, and $-36$ for $a_n \\$in the formula for the 4th term:}\)

\(\begin{array}{rcll} a_4 &=& a_1 + (n-1)d \qquad & \text{This gives the 4th term of any arithmetic sequence.} \\ -36 &=& -21 + (4-1)d \qquad & \text{Substitute.} \\ -36 &=& -21 + 3d \qquad & \text{Subtract within the parentheses.}\\ -15 &=& 3d \qquad & \text{Subtract -21 from both sides.} \\ -5 &=& d \qquad & \text{To isolate d, divide both sides by 3.} \\ \end{array}\)

\(\text{To find the two arithmetic means between $-21$ and $-36$, $\\$we add the common difference $-5$, as shown:}\)

\(\begin{array}{rcll} -21+d &=& -21 +(-5) \\ &=& -26 \qquad & \text{This is $a_2$.} \\ \end{array}\)

\(\begin{array}{rcll} -21+2d &=& -21 +2(-5) \\ &=& -21 -10 \\ &=& -31 \qquad & \text{This is $a_3$.} \\ \end{array}\)

\(\text{Two arithmetic means between $-21$ and $-36$ are $-26$ and $-31$.} \)

If x, y, and z are positive with xy = 20(cube root 2), xz = 35 (cube root 2), and yz = 14(cube root 2),

then what is xyz?

\(\begin{array}{|rcll|} \hline xy &=& 20\sqrt[3]{2} \\ xz &=& 35\sqrt[3]{2} \\ yz &=& 14\sqrt[3]{2} \\\\ xy\cdot xz\cdot yz &=& 20\sqrt[3]{2}\cdot 35\sqrt[3]{2} \cdot 14 \sqrt[3]{2} \\ x^2y^2z^2 &=& 20\cdot 35 \cdot 14 \cdot (\sqrt[3]{2} )^3 \\ (xyz)^2 &=& 20\cdot 35 \cdot 14 \cdot 2 \\ xyz &=& \sqrt{20\cdot 35 \cdot 14 \cdot 2} \\ xyz &=& \sqrt{19600} \\ \mathbf{xyz} &\mathbf{=} & \mathbf{140} \\ \hline \end{array}\)

The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$, then what are the $x$-intercepts of the graph of $f$?

The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$,

then what are the $x$-intercepts of the graph of $f$?

The polynomial of degree 3 is:

\(f(x) = ax^3+bx^2+cx+d\)

\(f(0) = 0:\)

\(\begin{array}{|rcll|} \hline f(0) = 0 &=& a\cdot 0^3+b\cdot 0^2+c\cdot 0+d \\ 0 &=& d \\ \mathbf{d} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

The polyomial of degree 3 is now:

\(f(x) = ax^3+bx^2+cx\)

\(f(-1) = 15:\)

\(\begin{array}{|rcll|} \hline f(-1) = 15 &=& a\cdot (-1)^3+b\cdot (-1)^2+c\cdot (-1) \\ 15 &=& -a +b-c \\ \mathbf{ -a+b-c} &\mathbf{=}& \mathbf{15} \qquad (1) \\ \hline \end{array}\)

\(f(1) = -5:\)

\(\begin{array}{|rcll|} \hline f(1) = -5 &=& a\cdot 1^3+b\cdot 1^2+c\cdot 1 \\ -5 &=& a +b+c \\ \mathbf{ a +b+c } &\mathbf{=}& \mathbf{-5} \qquad (2) \\ \hline \end{array}\)

\(f(2) = 12:\)

\(\begin{array}{|rcll|} \hline f(2) = 12 &=& a\cdot 2^3+b\cdot 2^2+c\cdot 2 \\ 12&=& 8a+4b+2c \quad & | \quad : 2 \\ \mathbf{ 4a +2b+c } &\mathbf{=}& \mathbf{6} \qquad (3) \\ \hline \end{array}\)

find a,b,c:

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{ -a+b-c} &\mathbf{=}& \mathbf{15} \\ (2) & \mathbf{ a +b+c } &\mathbf{=}& \mathbf{-5} \\ (3) & \mathbf{ 4a +2b+c } &\mathbf{=}& \mathbf{6} \\ \hline (1)+(2): & 2b &=& 10 \\ & \mathbf{b} &\mathbf{=}& \mathbf{5} \\ \hline (3)-(2): & 3a+b &=& 11 \quad & | \quad b=5 \\ & 3a+5 &=& 11 \\ & 3a &=& 6 \\ & \mathbf{a} &\mathbf{=}& \mathbf{2} \\ \hline (2) & a +b+c & = & -5 \quad & | \quad a=2\quad b=5 \\ & 2 +5+c & = & -5 \\ & 7+c & = & -5 \\ & c & = & -5 -7 \\ & \mathbf{c} &\mathbf{=}& \mathbf{-12} \\ \hline \end{array}\)

The polyomial of degree 3 is:

\(f(x) = 2x^3+5x^2-12x\)

The graph:

\(\text{what are the $x$-intercepts of the graph of $f\ $ ?}\)

\(\begin{array}{|lrcll|} \hline & 2x^3+5x^2-12x &=& 0 \\ & x(2x^2+5x-12) &=& 0 \\ \hline 1. & \mathbf{x_1} &\mathbf{=}& \mathbf{0} \\ \hline 2. \\ & 2x^2+5x-12 &=& 0 \\ & x &=& \dfrac{-5 \pm \sqrt{25-4\cdot 2 \cdot (-12) } } {2\cdot 2} \\ & &=& \dfrac{-5 \pm \sqrt{121} } {4} \\ & &=& \dfrac{-5 \pm 11 } {4} \\\\ & x_2 &=& \dfrac{-5 + 11 } {4} \\ & \mathbf{x_2} &\mathbf{=}& \mathbf{1.5} \\\\ & x_3 &=& \dfrac{-5 - 11 } {4} \\ & \mathbf{x_3} &\mathbf{=}& \mathbf{-4} \\ \hline \end{array}\)

\(\text{ The $x$-intercepts of the graph of $f$ are x=-4, x=0, and x = 1.5 }\)

What is the sum of the series?

1.

\(\begin{array}{|rcll|} \hline && \displaystyle \sum \limits_{i=1}^{5} 4i \\ &=& 4\cdot 1 + 4\cdot 2 + 4\cdot 3 + 4\cdot 4 + 4\cdot 5 \\ &=& 4+8+12+16+20 \\ &=& 60 \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline && \displaystyle \sum \limits_{k=1}^{4} (2k)^2 \\ &=& (2\cdot 1)^2 + (2\cdot 2)^2 + (2\cdot 3)^2 + (2\cdot 4)^2 \\ &=& 4+16+36+64 \\ &=& 120 \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline && \displaystyle \sum \limits_{k=3}^{6} (-5k+20) \\ &=& (-5\cdot 3+20) + (-5\cdot 4+20) + (-5\cdot 5+20) + (-5\cdot 6+20) \\ &=& 5+0-5-10 \\ &=& -10 \\ \hline \end{array}\)

Which answer represents the series in sigma notation?

1.
\(\begin{array}{|rcll|} \hline && 1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729} \\ &=& \sum \limits_{j=1}^{15} \dfrac{1}{3^{j-1}} \\ \hline \end{array} \)

2.

\(\begin{array}{|rcll|} \hline && -3+(-1)+1+3+5 \\ &=& \sum \limits_{j=1}^{5} (2j-5) \\ \hline \end{array}\)

So I’m starting to see that summation n = 0 to infinity of (a/b)^n *n = ab/(b-a)^2 , (where b>a so it actually converges )...? This is actually true for all cases, and a known result? If so, I’d like to know what you’d actually call it so I can look up the proof as it’s never been mentioned or taught at my Uni, although there’s been worked solutions using this... Thanks

\(\text{Let $x = \dfrac{a}{b} \quad b > a\quad $ or $\quad|x|<1$ }\)

Now we have:

\(\displaystyle \sum \limits_{n=0}^{\infty} nx^n\)

Let the progression to be summed be put equal to s:

\(s = 0+x+2x^2+3x^3+4x^4+\cdots + nx^n +\cdots\)

It is divided by x and multiplied by dx then

\(\dfrac{s\ dx}{x} = dx +2x\ dx+3x^2\ dx+4x^3\ dx+\cdots + nx^{n-1}\ dx +\cdots \)

and with the integrals taken this equation is found

\(\displaystyle \int \dfrac{s\ dx}{x} = x +x^2+x^3+x^4 +\cdots + x^n + \cdots = \dfrac{x}{1-x} \quad \text{ infinite geometric progression }\)

Therefore from the equation:

\(\displaystyle \int \dfrac{s\ dx}{x} = \dfrac{x}{1-x}\)

on differentiation s is found. For the equation becomes:

\(\displaystyle \dfrac{s}{x} = \dfrac{1}{1-x} + x(-1)(1-x)^{-2}(-1) \\ \displaystyle \dfrac{s}{x} = \dfrac{1}{1-x} + \dfrac{x}{(1-x)^2} \\ \displaystyle \dfrac{s}{x} = \dfrac{1-x+x}{(1-x)^2} \\ \displaystyle \dfrac{s}{x} = \dfrac{1 }{(1-x)^2} \)

thus there is produced:

\(\displaystyle s = \dfrac{x}{(1-x)^2} \)

So \(\text{let $x = \dfrac{a}{b} $}:\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{ \dfrac{a}{b} } {\left(1-\dfrac{a}{b} \right)} \\\\ &=& \dfrac{ a } {b \dfrac{(b-a)^2}{b^2} } \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{ ab } { (b-a)^2 }} \\ \hline \end{array}\)

Finally

\(\displaystyle \sum \limits_{n=0}^{\infty} n \left(\dfrac{a}{b}\right)^n = \dfrac{ ab } { (b-a)^2 } \)