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# Algebra Help

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

Dec 5, 2017

#1
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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)  =  $$\frac{1}{4^2}$$   =   $$\frac{1}{16}$$

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Dec 5, 2017

#1
+7347
+3

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