what happens to the shape of the graph of f(x)=b^x when b=1?
what happens to the shape of its inverse?
explain why this happens
To determine what happens to the shape of the graph of the function \( f(x) = b^x \) when \( b = 1 \), we need to substitute \( b = 1 \) into the equation.
First, let's consider the graph of \( f(x) = b^x \). When \( b = 1 \), the equation becomes \( f(x) = 1^x \). Any number raised to the power of 0 is 1, so \( 1^x = 1 \) for all values of \( x \).
Therefore, when \( b = 1 \), the graph of \( f(x) = 1^x \) is a horizontal line located at \( y = 1 \) for all values of \( x \).
Now, let's consider the inverse of \( f(x) = b^x \), denoted as \( f^{-1}(x) \). To find the inverse, we need to swap the roles of \( x \) and \( y \), and solve the equation for \( y \).
Starting with \( y = b^x \), we swap \( x \) and \( y \) to get \( x = b^y \). Next, we solve for \( y \) by taking the logarithm of both sides of the equation. Assuming \( b \) is positive and not equal to 1, we use the natural logarithm (base \( e \)) to solve for \( y \):
\( y = \log_b(x) \)
However, when \( b = 1 \), the equation becomes \( y = \log_1(x) \). This equation is undefined because there is no meaningful way to express \( 1 \) raised to what power equals \( x \). In other words, the logarithm with respect to \( b = 1 \) does not exist.
Hence, when \( b = 1 \), the shape of the inverse function \( f^{-1}(x) \) does not exist. It is not a valid function because its inverse is undefined.