The function can be defined as binary relation between two sets which associates each element of the frist set to exactly one element of the second set. Let's say that two sets, named A and B exists. Than the function is a process that associates each
element of set A, to a single element of a set B. Function from A to B is usually denoted with letters \(f,g,h,...\) Function \(f\) from A to B is mathematically decsribed as: $$ f: A\longrightarrow B, f(a) = b, a\in A, b\in B.$$ The function
\(f\) is even if \(f(-x) = f(x)\) for every \(x\) from domain. The graph of even function is symmetric with respect to y-axis. The function \(f\) is odd if \(f(-x)=-f(x)\) for every \(x\) from domain. The graph of odd function is symmetric
with respect to orgin of coordinate system. The function f increases on the interval \(x_1,x_2\) if for every two elements \(x_1,x_2\) form function domain f, for which \(x_1
< x_2\) is valid \(f(x_1) \leq f(x_2)\) and strongly increase if the relation \(f(x_1) < f(x_2)\) is valid. The value of function f decereases in range \(\langle x_1,x_2\rangle\) if for every two elements \(x_1, x_2\) from function domain \(f\), for which
\(x_1 < x_2\) the relationship \(f(x_1) \geq f(x_2)\) is valid, and strongly decreases if the relationship \(f(x_1)> f(x_2) \) is valid. Decreasing and increasing functions are both calld monotonic functions.
The function is constnat on interval \(\langle x_1, x_2 \rangle\) if every two elements \(x_1, x_2\) from function domain the relation \(f(x_1) = f(x_2).\)
The function is constnat on interval \(\langle x_1, x_2 \rangle\) if every two elements \(x_1, x_2\) from function domain the relation \(f(x_1) = f(x_2).\)
- Example 1 - Determine the value of the function \(f(x) = \frac{x+1}{x-1}\) if \(x = -1\) and \(x=1.\)
Solution For \(x = -1\) $$ f(-1) = \frac{-1+1}{-1-1} = -\frac{1}{2}.$$ For \(x = 1\) the value of the function \(f(x)\) does not exist i.e. $$ f(1) = \frac{1+1}{1-1} = \frac{2}{0}$$ - Example 2 - In Figures a,b,c, and d four different mathematical functions are shown. Which figure represents a function that only increases in range from 0 to 10 ? Which figure have repeating minimum and maximum values ?
a) b) c) d)
Solution - Figure a is increasing in range from 0 to 10. Figure d has repeating minimum and maximum values. - Example 3 - The function \(f(x) = \sin(x)\) is shown in following figure. How would absoulte value of that function look like.
Solution - Example 4 - Two functions are give \(f(x) = x+2 \), and \(g(x) = |x+3|\). Which one is greater \(f(-5)\) or \(g(-5)\) ?
Solution $$ f(-5) = -5 + 2 = -3$$ $$ g(-5) = |-5+3| = |-2| = 2$$ $$ g(-5) > f(-5)$$ - Example 5 - Determine if function \(f(x) = x^2 \) is even or odd ?
Solution To investigate if the function is even or odd lets select two x values for example -3 and 3. For \(x=-3\): $$f(-3) = (-3)^2 = 9, $$ and for \(x=3\): $$ f(3) = 3^2 = 9.$$ From the obtained vlaues it can be concluded that the function is even i.e. $$ f(-3) = 9, f(3) = 9 \Rightarrow f(-3) = f(3).$$ - Example 6 - Determine if function \(f(x) = x\) is even or odd?
Solution - To investigate if the function is even or odd let's select two x values for example -4 and 4. for \(x=-4\): $$ f(-4) = -4, $$ and for \(x=4\): $$ f(4) = 4$$ From the obtained solutions it can be concluded that the function is odd i.e. $$ f(-4) = -4, f(4) = 4 \Rightarrow f(-4)=-f(4)$$ Function is odd. - Example 7 - Determine if the function \(f(x) = x^2-1\) is even or odd ?
SolutionAs in previous cases select two values of x (positive and negative) i.e. -5 and 5. For \(x=-5\): $$f(-5) = (-5)^2 -1 = 25 -1 = 24, $$ and for \(x = 5\): $$ f(5) = 5^2 -1 = 24.$$ The function is even i.e. \(f(-5) = f(5)\) - Example 8 - Determine if the function \(f(x) = \frac{x-1}{x+1}\) is even or odd function?
Solution Investigate what would happen if argument x is negative $$ f(-x) = \frac{-x-1}{-x+1} = \frac{-(x+1)}{-(x-1)} = \frac{x+1}{x-1} \Rightarrow f(-x)\neq f(x).$$ The function is not even nor odd. - Example 9 - Determine if the function \(f(x) = x^3 - x\) is even or odd ?
Solution - $$f(-x) = -x^3 +x = -(x^3-x) = -f(x) $$ The function is odd. - Example 10 - Determine if function \(f(x) = 2x-3\) is increasining or decerasing function.
Solution To define if function is increasing or decreasing pick two \(x\) values i.e. 5 and 6 and calculate the f(5), and f(6) values. $$ f(5) = 2\cdot 5 -1 \Rightarrow f(5) = 9,$$ $$ f(6) = 2\cdot 6 - 1 = 11. $$ From previous calculations it can be concluded that the function \(f(x) = 2x-3\) is increasing function due to tha fact that \(f(5) < f(6).\) - Example 11 - If \(f(\sqrt{x}+1) = 3x+2\), determine the \(f(x)\) is equal to
- a) \(3x^2 + 6x -5\)
- b) \(3x^2- 6x -5\)
- c) \(3x^2 + 6x +5\)
- d) \(3x^2 - 6x +5\)
Solution a) Let's try by determining \(f(\sqrt{x}+1)\) of function \(f(x) = 3x^2 + 6x - 5.\) $$\begin{eqnarray}f(\sqrt{x}+1) &=& 3(\sqrt{x}+1)^2 + 6(\sqrt{x}+1) -5 \\ \nonumber f(\sqrt{x}+1) &=& 3x + 6\sqrt{x}+ 3 + 6\sqrt{x}+ 6 - 5 \\\nonumber f(\sqrt{x}+1) &=& 3x + 12 \sqrt{x} + 1\end{eqnarray}$$ From obtained solution it can be noticed that \(f(x) = 3x^2 + 6x - 5\) is not the solution.
b) Investigate if \(f(x) = 3x^2 -6x - 5\) is the original function. $$\begin{eqnarray}f(\sqrt{x}+1 &=& 3(\sqrt{x}+1)^2 -6(\sqrt{x}+1) -5\\\nonumber f(\sqrt{x}+1) &=& 3x+6\sqrt{x} + 3 - 6\sqrt{x} - 6 -5 \\ \nonumber f(\sqrt{x}+1) &=& 3x - 8\end{eqnarray}$$ From obtained solution it can be seen that the \(f(x) = 3x^2 -6x - 5\) is not the solution.
c) Investigate if \(f(x) = 3x^2 + 6 x +5 \) is the original function. $$\begin{eqnarray}f(\sqrt{x}+ 1) &=& 3(\sqrt{x}+1)^2 +6(\sqrt{x}+1) + 5\\ \nonumber f(\sqrt{x} + 1) &=& 3x + 6\sqrt{x} + 3 + 6\sqrt{x} + 6 + 5 \\\nonumber f(\sqrt{x}+1) &=& 3x + 12\sqrt{x} + 14\end{eqnarray}$$ From the obtained solution it can be concluded that c) is not the original function.
d) Investigate if \(f(x) = 3x^2 -6x + 5\) is the original function. $$\begin{eqnarray}f(\sqrt{x}+1) &=& 3(\sqrt{x}+1)^2 -6(\sqrt{x}+1) +5 \\ \nonumber f(\sqrt{x}+1) &=& 3x + 6\sqrt{x} + 3 -6\sqrt{x} - 6 + 5 \\ \nonumber f(\sqrt{x}+1) &=& 3x +2 \end{eqnarray}$$ From obtained solution it can be concluded that \(f(x) = 3x^2 -6x + 5\) is the original function. - Example 12 - If \(f(x) = 2x^2 +3x\) is the original function determine the value of \(f\left(\frac{1}{1+\sqrt{2}}\right).\)
Solution$$\begin{eqnarray}f\left(\frac{1}{1+\sqrt{2}}\right)&=&2\left(\frac{1}{1+\sqrt{2}}\right)^2 + 3\left(\frac{1}{1+\sqrt{2}}\right)\\ \nonumber f\left(\frac{1}{1+\sqrt{2}}\right)&=&\frac{2}{\left(1+\sqrt{2}\right)^2} + \frac{3}{1+\sqrt{2}} \\\nonumber f\left(\frac{1}{1+\sqrt{2}}\right)&=& \frac{2 + 3(1+\sqrt{2})}{\left(1+\sqrt{2}\right)^2}\\ \nonumber f\left(\frac{1}{1+\sqrt{2}}\right)&=& \frac{5 + 3\sqrt{2}}{1+2\sqrt{2} + 2}\\ \nonumber f\left(\frac{1}{1+\sqrt{2}}\right)&=& \frac{5 + 3\sqrt{2}}{3+2\sqrt{2}}\cdot \frac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ \nonumber f\left(\frac{1}{1+\sqrt{2}}\right)&=& \frac{15 - 10\sqrt{2} + 9\sqrt{2} - 12}{9-6\sqrt{2}+6\sqrt{2} - 8} \\ \nonumber f\left(\frac{1}{1+\sqrt{2}}\right)&=& 3-\sqrt{2}\end{eqnarray} $$ - Example 13 - If \(f(\sqrt{x}-1) = 16x - 2\sqrt{x} -15\) then \(f(x^2)\) is equal to
- \(-16x^4 + 30x^2 -1\)
- \(16x^4 + 30x^2 -1\)
- \(16x^4-30x^2-1\)
- \(16x^4+30x^2+1\)
Solution The procedure is to find \(f(x)\) from offered \(f(x^2)\) and then insert \(\sqrt{x}-1.\)
The \(f(x)\) of solution a. $$\begin{eqnarray}f(x)&=&-16x^2 +30x - 1\end{eqnarray}$$ Now change the function argument from \(x\) to \(\sqrt{x}-1.\) $$\begin{eqnarray}f(\sqrt{x}-1) &=& -16(\sqrt{x}-1)^2 + 30(\sqrt{x}-1) -1 \\\nonumber f(\sqrt{x}-1) &=& -16(x-2\sqrt{x} +1) + 30\sqrt{x}-30 -1\\\nonumber f(\sqrt{x}-1) &=& -16x+32\sqrt{x}-16 +30\sqrt{x}-30 -1 \\ \nonumber f(\sqrt{x}-1) &=& -16x+62\sqrt{x} - 47\end{eqnarray}$$
The \(f(x)\) of solution b can be written in the following form $$ f(x) = 16x^2 +30x -1$$ Now change the function argument from \(x\) to \(\sqrt{x}-1.\) $$\begin{eqnarray}f(\sqrt{x}-1) &=& 16(\sqrt{x}-1)^2 +30 (\sqrt{x}-1) - 1\\\nonumber f(\sqrt{x}-1) &=& 16x-32\sqrt{x} +16 +30\sqrt{x}-30 -1 \\\nonumber f(\sqrt{x}-1) &=& 16x -2\sqrt{x}-15\end{eqnarray}$$ - Example 14 - If \(f\left(\sqrt{x}\right) = 1+2\left(\sqrt{x}-x\right)\) then \(f(x)\) is equal to a)\(1-2x-2x^2,\) or b)\(1+2x-2x^2.\)
Solution - Let's assume the the original function \(f(x)\) is equal to \(1-2x-2x^2,\) then $$\begin{eqnarray}f(x) &=& 1-2x-x^2\\ \nonumber f(\sqrt{x})&=& 1 -2\sqrt{x} - 2x\\ \nonumber f(\sqrt{x})&=& 1 -2(\sqrt{x} +x)\end{eqnarray}$$ From obtained solution it can be concluded that from this function the \(f(\sqrt{x}) = 1+2(\sqrt{x}-x)\) cannot be obtained.
- Now let's assume that the original function \(f(x)\) is \(1+2x-2x^2,\) then $$ \begin{eqnarray}f(x) &=& 1+2x-2x^2 \\\nonumber f(\sqrt{x})&=& 1+2\sqrt{x}-2x\\ \nonumber f(\sqrt{x})&=&1+2(\sqrt{x}-x)\end{eqnarray}$$ From the obtained solution it can be concluded that the \(f(x)=1+2x-2x^2\) is the function from which the \(f(\sqrt{x}) = 1+2(x-x^2)\) can be obtained.
- Example 15 - If \(f(2x-1) = -4x^2+24x-10\) then f(x) is equal to a) \(-x^2+10x-1,\) b) \(-x^2+10x+1,\) c) \(-x^2-10x-1\) or d)\(-x^2-10x+1.\)
Solution - The procedure is to test each function given by inserting the argument \(2x-1\) instead of \(x\).- $$\begin{eqnarray}f(x)&=& -x^2+10x-1\\ \nonumber f(2x-1) &=& -(2x-1)^2 +10(2x-1)-1\\ \nonumber f(2x-1)&=& -(4x^2-4x+1) +20x-10 -1 \\ \nonumber f(2x-1)&=& -4x^2 +4x-1 +20x-10 -1\\\nonumber f(2x-1)&=&-4x^2 + 24x-12\end{eqnarray}$$ The f(x) is not \(-x^2+10x-1.\)
- $$\begin{eqnarray}f(x)&=&-x^2+10x+1\\\nonumber f(2x-1)&=& -(2x-1)^2 +10(2x-1) +1 \\\nonumber f(2x-1) &=& -(4x^2-4x+1) + 20x-10 +1 \\\nonumber f(2x-1)&=& -4x^2 +4x -1 +20x-10 +1 \\\nonumber f(2x-1)&=&-4x^2+24x-10\end{eqnarray}$$ The f(x) is \(-x^2+10x+1.\)
- $$\begin{eqnarray}f(x) &=& -x^2 -10x-1 \\\nonumber f(2x-1)&=& -(2x-1)^2 -10(2x-1)-1\\ \nonumber f(2x-1)&=& -4x^2+4x-1 -20x+10 -1 \\ \nonumber f(2x-1)&=& -4x^2-16x+8 \end{eqnarray}$$ The f(x) is not \(-x^2-10x-1.\)
- $$\begin{eqnarray}f(x)&=& -x^2-10x-1\\ \nonumber f(2x-1)&=& -(2x-1)^2 -10(2x-1)-1\\ \nonumber f(2x-1)&=& -(4x^2-4x+1)-20x+10 -1 \\\nonumber f(2x-1) &=& -4x^2+4x-1 -20x+10-1 \\\nonumber f(2x-1) &=& -4x^2 +4x-1 -20x+10 -1\\ \nonumber f(2x-1) &=& -4x^2-16x+8\end{eqnarray}$$ The f(x) is not \(-x^2-10x+1.\)
- Example 16 - From functions shown in following figures determine which one is an even function?
a) b) c) d)
Solution As already stated the function is even if the relation \(f(-x) = f(x)\) is valid. This means that for positive and negative values of x (for example -2 and 2) the output value of functions i.e. \(f(x)\) or \(f(-x)\) are equal. In previously shown figures the only function that could satisfy this relationship is function shown in figure d). - Example 17 - Calculate the value of the function \(f(x)=\frac{x^2+4}{x^2-4},\) for \(x_1 = 5,\) \(x_2 = -2,\) \(x_3 = 2,\) \(x_4 = \frac{1}{t},\) \(x_5 = \frac{1}{t-1}. \)
SolutionFor \(x_1 = 5\) the value of \(f(5)\) is $$\begin{eqnarray}f(5)&=& \frac{5^2+4}{5^2-4}\\ \nonumber f(5)&=& \frac{25+4}{25-4} \\ \nonumber f(5) &=& \frac{29}{21}\end{eqnarray}$$ For \(x_2= -2\) the value of \(f(-2)\) $$\begin{eqnarray}f(-2)&=& \frac{(-2)^2+4}{(-2)^2-4} \\ \nonumber f(-2)&=& \frac{4+4}{4-4} \\ \nonumber f(-2)&=& \frac{8}{0}.\end{eqnarray}$$ The solution does not exist.
For \(x_3 =2\) the value of \(f(2)\) $$\begin{eqnarray}f(2)&=& \frac{4+4}{4-4}\\\nonumber f(2) &=& \frac{8}{0}\end{eqnarray}$$ The solution does not exist.
For \(x_4 = \frac{1}{t}\) the value of \(f(\frac{1}{t})\) $$\begin{eqnarray}f\left(\frac{1}{t}\right)&=&\frac{\frac{1}{t^2}+4}{\frac{1}{t^2}-4}\\\nonumber f\left(\frac{1}{t}\right)&=& \frac{\frac{t^2+4}{t^2}}{\frac{t^2-4}{t^2}}\\\nonumber f\left(\frac{1}{t}\right)&=& \frac{t^2+4}{t^2-4}.\end{eqnarray}$$ For \(x_5=\frac{1}{t-1}\) the value of \(f\left(\frac{1}{t-1}\right)\) $$\begin{eqnarray}f\left(\frac{1}{t-1}\right)&=&\frac{\frac{1}{(t-1)^2}+4}{\frac{1}{(t-1)^2}-4}\\\nonumber f\left(\frac{1}{t-1}\right)&=&\frac{\frac{1+4t^2-8t+4}{t^2-2t+1}}{\frac{1-4t^2+8t-4}{t^2-2t+1}}\\\nonumber f\left(\frac{1}{t-1}\right)&=&\frac{4t^2-8t+5}{-4t^2+8t-3}\end{eqnarray}$$ - Example 18 - From functions shown in following figures determine which one is an odd function?
a) b) c) d)
Solution - The function is odd when the relation \(f(-x)=-f(x)\) is valid. From functions shown in previous figures it can be seen that only the function a) satisfy this relationship i.e. the function is odd. For x vlaues 1 and -1 the output values of the function is 1 and -1, respectively. - Example 19 - Calculate the value of function \(f(x) = 2^{\frac{x-1}{2}}+1\) if variable x is equal to 1, 0, 3, -9, and \(1+2\log_2(x-1).\)
Solution $$\begin{eqnarray}f(1) &=& 2^{\frac{1-1}{2}}+1\\ \nonumber f(1) &=& 1+1 = 2\end{eqnarray}$$ $$\begin{eqnarray}f(0) &=& 2^{\frac{0-1}{2}}+1\\\nonumber f(0) &=& \frac{1}{\sqrt{2}} + 1 \\\nonumber f(0) &=& \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} +1 \\\nonumber f(0)&=&\frac{\sqrt{2}}{2}+1\end{eqnarray}$$ $$\begin{eqnarray}f(3) &=& 2^{\frac{3-1}{2}}+1\\ \nonumber f(3) &=& 2^{\frac{2}{2}}+1\\\nonumber f(3) &=& 3\end{eqnarray}$$ $$\begin{eqnarray}f(-9) &=& 2^{\frac{-9-1}{2}} + 1 \\\nonumber f(-9) &=& 2^{-5}+1 \\ \nonumber f(-9) &=& \frac{33}{32}\end{eqnarray}$$ $$\begin{eqnarray}f(1+2\log_2(x-1))&=& 2^{\frac{1+2\log_2(x-1)-1}{2}} +1 \\ \nonumber f(1+2\log_2(x-1))&=&2^{\log_2(x-1)}+1\\ \nonumber f(1+2\log_2(x-1))&=& x\end{eqnarray}$$ $$\begin{eqnarray}\end{eqnarray}$$ - Example 20 - Four different functions are shown in following figures. Determine which function is the increasing function.
a) b) c) d)
Solution By definition, a function \(f\) increases on interval \(\langle x_1, x_2\rangle\) if for every two elements \(x_1,x_2\) from function domain, for which \(x_1 < x_2\) the relation \(f(x_1) \leq f(x_2)\), and strongly increases if the relation \(f(x_1) < f(x_2)\) is valid. Taking into account this definition and applying it to the previous figures, it can be concluded that the function b is a increasing function. - Example 21 - If \(f(x) =\frac{2x+1}{2x-1}\) solve the inequality \(f(x+1) > 1.\)
Solution $$\begin{eqnarray}f(x+1)&=&\frac{2(x+1)+1}{2(x+1)-1}\\\nonumber f(x+1)&=&\frac{2x+2+1}{2x+2-1}\\\nonumber f(x+1)&=&\frac{2x+3}{2x+1}\end{eqnarray}$$ $$\begin{eqnarray}\frac{2x+3}{2x+1} > 1\\ \nonumber 2x+1>0 \\\nonumber x > -\frac{1}{2}\end{eqnarray}$$ - Example 22 - Four different functions are shown in following figures. Determine which function is the decreasing.
a) b) c) d)
Solution - By definition, a function \(f\) decreases on interval \(\langle x_1, x_2\rangle\) if for every two elements \(x_1,x_2\) from function domain, for which \(x_1 < x_2\) the relation \(f(x_1) \geq f(x_2)\), and strongly decreases if the relation \(f(x_1)> f(x_2)\) is valid. Taking into account this definition and applying it to the previous figures, it can be concluded that the function a is a decreasing function. - Example 23 - If \(f(x) = 5\cdot 3^x,\) for which \(x\) the relation \(\frac{x+1}{f(1-x)} > 0.\)
Solution$$\begin{eqnarray} f(x+1) = 5\cdot 3^{x+1}\end{eqnarray}$$ $$ f(1-x) = 5\cdot 3^{1-x}$$ $$\begin{eqnarray}\frac{5\cdot 3^{1+x}}{5\cdot 3^{1-x}} &>& 0 \\ \nonumber \frac{3\cdot 3^{x}}{3\cdot 3^{-x}} &>& 0 \\ \nonumber 3^2x &>& 0, x\in \mathbb{R}\end{eqnarray}$$ - Example 24 - Function is shown on following figure. Determine on which interval the function is increasing?
Solution As seen from figure the function is decreasing for x values in range \(-\infty\) to 0 and is increasing from 0 to \(+\infty\). The function is increasing on interval \(x \in [0, +\infty\rangle\). - Example 25 - Determine if the function \(f(x) = \frac{x^2-1}{x^5-x^3-x}\) is even or odd.
Solution $$\begin{eqnarray}f(-x)&=& \frac{(-x)^2-1}{(-x)^5-(-x)^3-(-x)}\\\nonumber f(-x)&=& \frac{x^2-1}{-x^5+x^3+x}\\ \nonumber f(-x)&=&\frac{x^2-1}{-(x^5-x^3-x)}\\ \nonumber f(-x) &=& -\frac{x^2-1}{x^5-x^3-x}\\\nonumber f(-x) &=& -f(x)\end{eqnarray}$$ The function is odd. - Example 26 - Determine if the function \(f(x) =|2x|-3\) is even or odd.
Solution$$\begin{eqnarray}f(-x) &=& |2(-x)| -3 \\ \nonumber f(-x)&=& |2x| -3\\\nonumber f(-x) &=& f(x)\end{eqnarray}$$ The function is even. - Example 27 - Determine if the function \(f(x) = \frac{2^x - 2^{-x}}{2^x+2^{-x}}\) is even or odd.
Solution$$\begin{eqnarray}f(-x)&=&\frac{2^{-x}-2^{-(-x)}}{2^{-x}+2^{-(-x)}}\\\nonumber f(-x) &=& \frac{2^{-x}-2^{x}}{2^{-x}+2^{x}}\\ \nonumber f(-x)&=&-\frac{2^{x}-2^{-x}}{2^{x} + 2^{-x}}\\ \nonumber f(-x)&=& -f(x)\end{eqnarray}$$ The function is odd. - Example 28 - Determine if the function \(f(x) = \frac{2x^2+3}{x^3+x^2}\) is even or odd.
Solution $$\begin{eqnarray}f(-x) &=& \frac{2(-x)^2+3}{(-x)^3 + (-x)^2}\\ \nonumber f(-x)&=&\frac{2x+3}{-x^3+x^2}\\\nonumber f(-x)&=& -\frac{2x+3}{x^3-x^2}\\\nonumber f(-x) &\neq& -f(x) \\\nonumber f(-x)&\neq&f(x)\end{eqnarray}$$ Function is not even nor odd. - Example 29 - If \(f(1-\sqrt{x}) = 4-\sqrt{x}-x,\) determine determine \(f\left(\frac{x}{2}\right)\)
Solution To determine \(f\left(\frac{x}{2}\right)\) first the \(f(x)\) must be determined. With little investigation we find that \(f(x)\) is equal to $$f(x) = 2+3x-x^2.$$ The proof for this is done by substituting \(1-\sqrt{x}\) in the original function. $$\begin{eqnarray}f(1-\sqrt{x})&=& 2+3(1-\sqrt{x})-(1-\sqrt{x})^2\\\nonumber f(1-\sqrt{x})&=&2+3-3\sqrt{x}-1+2\sqrt{x}-x \\\nonumber f(1-\sqrt{x})&=&4-\sqrt{x}-x\end{eqnarray}$$ Now substitute \(\frac{x}{2}\) in the original function. $$\begin{eqnarray}f\left(\frac{x}{2}\right) &=& 2+3\left(\frac{x}{2}\right) - \left(\frac{x}{2}\right)^2\\\nonumber f\left(\frac{x}{2}\right) &=& 2+\frac{3x}{2} - \frac{x^2}{4}\end{eqnarray}$$
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