Summarize the pertinont information obtained by applying the graptung stratogy and skotch the graph of $$ \mathrm{y}=\mathrm{I}(\mathrm{x}) $$. $$ f(\mathrm{x})=e^{0.3 x}+256 \mathrm{e}^{-0.5 \mathrm{x}} $$ Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The domain of I is (Type your answer in interval notation. Use a comma lo separate answers as needed.) B. The domain of I is empty.

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We begin with the function

$$
f(x)=e^{0.3x}+256\,e^{-0.5x}.
$$

A common graphing strategy is to look at the overall behavior of $$ f(x) $$ and then “reflect” the important information to the graph of the inverse function, which we denote by $$ I(x) $$. (Recall that if $$ f $$ is one‐to‐one on a chosen interval then its inverse has domain equal to the range of $$ f $$ on that interval.)

Since

• $$ e^{0.3x}>0 $$ for all $$ x\in\mathbb{R} $$ and  
• $$ 256\,e^{-0.5 x}>0 $$ for all $$ x\in\mathbb{R} $$,

we have $$ f(x)>0 $$ for every $$ x $$. In fact, as it turns out, $$ f(x) $$ has a unique minimum and then increases on either side. (Indeed, as $$ x\to\infty $$ the first term dominates, and as $$ x\to -\infty $$ the second term increases without bound.)

Thus, if we choose one of the monotonic pieces of $$ f(x) $$ (for example, the branch for $$ x\ge x_{\min} $$, where the minimum is attained) then $$ f $$ is one‐to‐one on that branch. In that situation, the inverse function $$ I(x) $$ is defined for

$$
x \ge f(x_{\min}).
$$

In other words, the domain of $$ I $$ is the range of $$ f $$ on that branch, which is

$$
\mathrm{dom}\,I=\Bigl[ f\bigl(x_{\min}\bigr),\infty\Bigr).
$$

To find the minimum value $$ f(x_{\min}) $$ we first compute the derivative:

$$
f'(x)=0.3\,e^{0.3x}-128\,e^{-0.5x}.
$$

Setting $$ f'(x)=0 $$ we obtain

$$
0.3\,e^{0.3x}=128\,e^{-0.5x}.
$$

Multiplying both sides by $$ e^{0.5x} $$ gives

$$
0.3\,e^{0.8x}=128,
$$

so

$$
e^{0.8x}=\frac{128}{0.3}.
$$

Taking natural log,

$$
0.8x=\ln\Bigl(\frac{128}{0.3}\Bigr)
$$

and hence

$$
x_{\min}=\frac{1}{0.8}\ln\Bigl(\frac{128}{0.3}\Bigr).
$$

Now, let

$$
A=e^{0.3x_{\min}}.
$$

Then, from the equation we found when setting the derivative equal to zero, we also have

$$
0.3\,A=128\, e^{-0.5x_{\min}}
\quad\Longrightarrow\quad e^{-0.5x_{\min}}=\frac{0.3\,A}{128}.
$$

Thus, the minimum value is

$$
f\bigl(x_{\min}\bigr)=A+256\,e^{-0.5x_{\min}}
=A+256\left(\frac{0.3\,A}{128}\right)
=A+ \left(\frac{256\times0.3}{128}\right)A.
$$

Notice that

$$
\frac{256\times0.3}{128}=0.6.
$$

So

$$
f\bigl(x_{\min}\bigr)=A+0.6\,A=1.6\,A.
$$

Recalling that

$$
A=e^{0.3x_{\min}}=\left(\frac{128}{0.3}\right)^{\frac{0.3}{0.8}}=\left(\frac{128}{0.3}\right)^{\frac{3}{8}},
$$

we deduce

$$
f\bigl(x_{\min}\bigr)=1.6\left(\frac{128}{0.3}\right)^{\frac{3}{8}}.
$$

Thus the range of $$ f $$ (when restricted to one branch where it is one‐to‐one) is

$$
\left[1.6\left(\frac{128}{0.3}\right)^{\frac{3}{8}}, \infty\right).
$$

Since the inverse function $$ I $$ has domain equal to the range of $$ f $$, we conclude that the correct answer is:

A. The domain of $$ I $$ is

$$
\boxed{\left[1.6\left(\frac{128}{0.3}\right)^{\frac{3}{8}}, \infty\right)}.
$$
A. The domain of $$ I $$ is $$\left[1.6\left(\frac{128}{0.3}\right)^{\frac{3}{8}}, \infty\right)$$.

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