Mathematics, 21.01.2020 01:31 ghetauto

F(x) = b^x and g(x) = log b x are inverse functions. explain why each of the following are true.

1. a translation of function f is f1(x) = b^(x-h). it is equivalent to a vertical stretch or vertical compression of function f.

2. the inverse of f1(x) = b^(x-h) is not equivalent to a translation of g.

3. the inverse of f1 (x) =b^(x-h) is not equivalent to a vertical stretch or vertical compression of g.

4. the function h(x) = log c x is a vertical stretch or compression of g or of its reflection -g. read this as"negative g"

will probably needs to use the properties of exponents and logarithms and change of base formulas to change the functions into alternate forms

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1. A translation of function f is $f_{1}(x)=b^{x-h}$ . It is equivalent to a vertical stretch or vertical compression of function f.

Let's take the function:

$f_{1}(x)=b^{x-h}$

If we used the rules of powers, we know that when we have a subtraction in the power of an exponential, it can be split into a division. So the function can be rewritten as:

$f_{1}(x)=\frac{b^{x}}{b^{h}}$

Remember the original function was:

$f(x)=b^{x}$

therefore:

$f_{1}(x)=\frac{f(x)}{b^{h}}$

this means that if $b^{h}$ then it will be a vertical stretch.

If $b^{h}1$ then it will be a vertical stretch.

2. The inverse of $f_{1}(x)=b^{x-h}$ is not equivalent to a translation of g.

This is partially true and you'll see why. Let's start by finding the inverse of that function:

$x=b^{y-h}$

we start by turning the given power to a multiplication of powers so we get:

$x=b^{-h}b^{y}$

we then move the $b^{-h}$ to the other side of the equation so we get:

$xb^{h}=b^{y}$

and turn the equation into a logarithm:

$y=log_{b}(xb^{h})$

so:

$g_{1}(x)=log_{b}(xb^{h})$

or:

$g_{1}(x)=g(b^{h}x)$

remember that when you multiply a constant by x, you will get a horizontal compression if $b^{h}1$ and a horizontal stretch if $b^{h}$ .

but there is another interpretation for this function. Let's take the original equation:

$x=b^{y-h}$

if we directly turned this equation into a logarithm we would get that:

$y-h=log_{b}(x)$

so:

$y=h+log_{b}(x)$

or:

$g_{1}(x)=h+g(x)$

if the inverse function is written like this, it can be interpreted as a vertical shift. Both interpretations are correct.

3. The inverse of $f_{1}(x)=b^{x-h}$ is not equivalent to a vertical stretch or vertical compression of g.

As we saw in the previous part of the problem, that function is either a horizontal stretch/compression or a vertical shift, not a vertical stretch or compression.

4. The function $h(x)=log_{c}(x)$ is a vertical stretch or compression of g or of its reflection -g.

We can rewrite the function like this thanks to log rules:

$h(x)=\frac{log_{b}(x)}{log_{b}(c)}$

which is the same as:

$h(x)=\frac{g(x)}{log_{b}(c)}$

If $log_{b}(c)1$ it will be a vertical compression of g(x). If $log_{b}(c)$ , then it will be a vertical stretch no matter if g is positive or negative.