Given:
The function is:
[tex]f(x)=2x^2-14x+10[/tex]
The function g(x), the image of f(x) after a [tex]r_{y=x}[/tex] (reflection over the line y=x).
To find:
The function g(x).
Solution:
We have,
[tex]f(x)=2x^2-14x+10[/tex]
Substitute [tex]f(x)=y[/tex] in the given function.
[tex]y=2x^2-14x+10[/tex]
The function g(x), the image of f(x) after a [tex]r_{y=x}[/tex] (reflection over the line y=x). So, interchange x and y.
[tex]x=2y^2-14y+10[/tex]
Now, we need to find the value of y.
[tex]x=2(y^2-7y+5)[/tex]
[tex]\dfrac{x}{2}=y^2-7y+5[/tex] [Divide both sides by 2]
[tex]\dfrac{x}{2}-5=y^2-7y[/tex] [Subtract 5 from both sides]
Add both sides half of square of coefficient of y, i.e. [tex](\dfrac{-7}{2})^2[/tex], to make it perfect square.
[tex]\dfrac{x}{2}-5+(\dfrac{-7}{2})^2=y^2-7y+(\dfrac{-7}{2})^2[/tex]
[tex]\dfrac{x}{2}-5+\dfrac{49}{4}=y^2-7y+(\dfrac{7}{2})^2[/tex]
[tex]\dfrac{x}{2}-5+\dfrac{49}{4}=\left(y-\dfrac{7}{2}\right)^2[/tex] [tex][\because (a-b)^2=a^2-2ab+b^2][/tex]
[tex]\dfrac{x}{2}+\dfrac{49-20}{4}=\left(y-\dfrac{7}{2}\right)^2[/tex]
Taking square root on both sides, we get
[tex]\pm\sqrt{\dfrac{x}{2}+\dfrac{29}{4}}=y-\dfrac{7}{2}[/tex]
[tex]\dfrac{7}{2}\pm\sqrt{\dfrac{2x+29}{4}}=y[/tex]
[tex]\dfrac{7}{2}\pm\dfrac{\sqrt{2x+29}}{2}=y[/tex]
[tex]\dfrac{7\pm \sqrt{2x+29}}{2}=y[/tex]
Substituting [tex]y=g(x)[/tex], we get
[tex]\dfrac{7\pm \sqrt{2x+29}}{2}=g(x)[/tex]
Therefore, the required function is [tex]g(x)=\dfrac{7\pm \sqrt{2x+29}}{2}[/tex].
