Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 7.1
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Exercise 97 Page 340

Practice makes perfect
a

Before we can find the inverse of the given function, we need to rewrite it as an equation relating x and y.

f(x)=sqrt(4x-1) ⇔ y=sqrt(4x-1) Now, to algebraically determine the inverse of the given equation we exchange x and y and solve for y. Given Equation & Inverse Equation y=sqrt(4 x-1) & x=sqrt(4 y-1) The result of isolating y in the new equation will be the inverse of the given function.

x=sqrt(4y-1)
x^3=(sqrt(4y-1))^3

( sqrt(a) )^n = a

x^3=4y-1
x^3+1=4y
x^3+1/4=y
y=x^3+1/4

Now that we have found y, we know the inverse of the given function. f^(- 1)(x)=x^3+1/4

b

Similarly as in Part A, to find the inverse of a function we start by exchanging the variables in the given function. Then we can solve for y.

ccc Given Function & & Inverse Function y=log_7 x & & x=log_7 yNow we have to isolate y in the inverse equation. The inverse of a logarithmic function is an exponential function. Therefore, we can use the definition of a logarithm to rewrite the equation. Definition:& x=log_b y &&⇔ b^x= y Equation:& x=log_7 y &&⇔ 7^x= y We found that y=7^x. This is the inverse function. g^(-1)(x) = 7^x