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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 $h^{-1}(4)$.

michaelcai  Dec 5, 2017

Best Answer 

 #1
avatar+7164 
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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)  =  \(\frac{1}{4^2}\)   =   \(\frac{1}{16}\)

hectictar  Dec 5, 2017
 #1
avatar+7164 
+3
Best Answer

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)  =  \(\frac{1}{4^2}\)   =   \(\frac{1}{16}\)

hectictar  Dec 5, 2017

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