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Compute
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Re: Compute
by Eigenvalue » Sat Sep 26, 2026 12:23 am
Let T(x)=y
y=[tex]\frac{x²+x}{x}[/tex]
To find the inverse function, switch the x and y values and solve for y in terms of x
x=[tex]\frac{y²+y}{y}[/tex]
x=y+1, y≠0
y=x-1, as y≠0, x≠1
y=[tex]\frac{x²+x}{x}[/tex]
To find the inverse function, switch the x and y values and solve for y in terms of x
x=[tex]\frac{y²+y}{y}[/tex]
x=y+1, y≠0
y=x-1, as y≠0, x≠1
- Eigenvalue
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Re: Compute
by nycmath » Sat Sep 26, 2026 10:55 am
Eigenvalue wrote:Let T(x)=y
y=[tex]\frac{x²+x}{x}[/tex]
To find the inverse function, switch the x and y values and solve for y in terms of x
x=[tex]\frac{y²+y}{y}[/tex]
x=y+1, y≠0
y=x-1, as y≠0, x≠1
Clean an sweet.
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