+619 a) Suppose \(f(x)=\frac 95x-4\). Does $f$ have an inverse? If so, find $f^{-1}(20)$.
b) Suppose \(g(x)=4x^2+8x+13\). Does $g$ have an inverse? If so, find $g^{-1}(25)$
c) Suppose \(h(x)=\frac{1}{\sqrt x}\) for $x>0$. Does $h$ have an inverse? If so, find $h^{-1}(4)$.
+9481 a) f(x) = \(\frac95\)x - 4 This has an inverse because it is just a linear equation.
y = \(\frac95\)x - 4 To find the inverse, solve this equation for x , so add 4 to both sides.
y + 4 = \(\frac95\)x Multiply both sides by \(\frac59\) .
\(\frac59\)(y + 4) = x So the inverse function is...
f-1(x) = \(\frac59\)(x + 4) And to find f-1(20) , plug in 20 for x into this function.
f-1(20) = \(\frac59\)(20 + 4) = \(\frac59\)(24) = \(\frac{40}3\)
b) g(x) = 4x2 + 8x + 13
g(x) does not have an inverse function because it would have two different y values for an x value, and for an equaton to qualify as a function, there can only be one y value for every x value.
c) h(x) = \(\frac1{\sqrt{x}}\) for x > 0 Yes this has an inverse.
y = \(\frac1{\sqrt{x}}\) To find the inverse, solve this equation for x .
y\(\sqrt{x}\) = 1
\(\sqrt{x}\) = \(\frac1{y}\) Square both sides.
x = \(\frac1{y^2}\) So the inverse function is..
f-1(x) = \(\frac1{x^2}\) for x > 0
f-1(4) = \(\frac{1}{4^2}\) = \(\frac{1}{16}\)
+9481 a) f(x) = \(\frac95\)x - 4 This has an inverse because it is just a linear equation.
y = \(\frac95\)x - 4 To find the inverse, solve this equation for x , so add 4 to both sides.
y + 4 = \(\frac95\)x Multiply both sides by \(\frac59\) .
\(\frac59\)(y + 4) = x So the inverse function is...
f-1(x) = \(\frac59\)(x + 4) And to find f-1(20) , plug in 20 for x into this function.
f-1(20) = \(\frac59\)(20 + 4) = \(\frac59\)(24) = \(\frac{40}3\)
b) g(x) = 4x2 + 8x + 13
g(x) does not have an inverse function because it would have two different y values for an x value, and for an equaton to qualify as a function, there can only be one y value for every x value.
c) h(x) = \(\frac1{\sqrt{x}}\) for x > 0 Yes this has an inverse.
y = \(\frac1{\sqrt{x}}\) To find the inverse, solve this equation for x .
y\(\sqrt{x}\) = 1
\(\sqrt{x}\) = \(\frac1{y}\) Square both sides.
x = \(\frac1{y^2}\) So the inverse function is..
f-1(x) = \(\frac1{x^2}\) for x > 0
f-1(4) = \(\frac{1}{4^2}\) = \(\frac{1}{16}\)
