The function $$ f(x)=6 x+7 x^{-1} $$ has one local minimum and one local maximum. This function has a local maximum at $$ x= $$ with value and a local minimum at $$ x= $$ with value Question Help:

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The first step is to differentiate the function
$$
f(x)=6x+7x^{-1}
$$
with respect to $$ x $$. Differentiating term‐by‐term, we have
$$
f'(x)=6-7x^{-2}.
$$

Next, we set the derivative equal to zero to find the critical points:
$$
6-7x^{-2}=0 \quad\Longrightarrow\quad 6=\frac{7}{x^2}.
$$
Multiplying both sides by $$ x^2 $$ gives
$$
6x^2=7,
$$
so
$$
x^2=\frac{7}{6}.
$$
Taking the square root yields two solutions:
$$
x=\sqrt{\frac{7}{6}} \quad \text{and} \quad x=-\sqrt{\frac{7}{6}}.
$$

To determine which is a maximum and which is a minimum, we compute the second derivative:
$$
f''(x)=\frac{d}{dx}\left(6-7x^{-2}\right)=14x^{-3}.
$$

Evaluating at $$ x=\sqrt{\frac{7}{6}} $$ (which is positive) gives:
$$
f''\left(\sqrt{\frac{7}{6}}\right)=\frac{14}{\left(\sqrt{\frac{7}{6}}\right)^3}.
$$
Since the denominator is positive, $$ f''\left(\sqrt{\frac{7}{6}}\right) > 0 $$, indicating a local minimum at
$$
x=\sqrt{\frac{7}{6}}.
$$

Evaluating at $$ x=-\sqrt{\frac{7}{6}} $$ gives:
$$
f''\left(-\sqrt{\frac{7}{6}}\right)=\frac{14}{\left(-\sqrt{\frac{7}{6}}\right)^3},
$$
and because the cube of a negative number is negative, we have $$ f''\left(-\sqrt{\frac{7}{6}}\right) < 0 $$, indicating a local maximum at
$$
x=-\sqrt{\frac{7}{6}}.
$$

Now we calculate the function values at these points.

For $$ x=\sqrt{\frac{7}{6}} $$:
$$
f\left(\sqrt{\frac{7}{6}}\right)=6\sqrt{\frac{7}{6}}+7\left(\sqrt{\frac{7}{6}}\right)^{-1}.
$$
Notice that
$$
6\sqrt{\frac{7}{6}}=6\cdot\frac{\sqrt{7}}{\sqrt{6}}=\frac{6\sqrt{7}}{\sqrt{6}}
$$
and
$$
7\left(\sqrt{\frac{7}{6}}\right)^{-1}=7\cdot\frac{\sqrt{6}}{\sqrt{7}}=\frac{7\sqrt{6}}{\sqrt{7}}.
$$
A useful observation is that both terms simplify to $$ \sqrt{42} $$. To see this, note:
$$
6\sqrt{\frac{7}{6}}=\sqrt{6^2\cdot\frac{7}{6}}=\sqrt{6\cdot7}=\sqrt{42},
$$
and similarly,
$$
7\left(\sqrt{\frac{7}{6}}\right)^{-1}=\sqrt{7^2\cdot\frac{6}{7}}=\sqrt{7\cdot6}=\sqrt{42}.
$$
Thus,
$$
f\left(\sqrt{\frac{7}{6}}\right)=\sqrt{42}+\sqrt{42}=2\sqrt{42}.
$$

For $$ x=-\sqrt{\frac{7}{6}} $$:
$$
f\left(-\sqrt{\frac{7}{6}}\right)=6\left(-\sqrt{\frac{7}{6}}\right)+7\left(-\sqrt{\frac{7}{6}}\right)^{-1}.
$$
This simplifies to:
$$
f\left(-\sqrt{\frac{7}{6}}\right)=-6\sqrt{\frac{7}{6}}-7\left(\sqrt{\frac{7}{6}}\right)^{-1}.
$$
Using the same reasoning as above,
$$
-6\sqrt{\frac{7}{6}}=-\sqrt{42} \quad \text{and} \quad -7\left(\sqrt{\frac{7}{6}}\right)^{-1}=-\sqrt{42},
$$
so
$$
f\left(-\sqrt{\frac{7}{6}}\right)=-\sqrt{42}-\sqrt{42}=-2\sqrt{42}.
$$

Thus, the function has:

- A local maximum at $$ x=-\sqrt{\frac{7}{6}} $$ with value $$ -2\sqrt{42} $$,
- A local minimum at $$ x=\sqrt{\frac{7}{6}} $$ with value $$ 2\sqrt{42} $$.
The function $$ f(x) = 6x + 7x^{-1} $$ has: - A local maximum at $$ x = -\sqrt{\frac{7}{6}} $$ with value $$ -2\sqrt{42} $$, - A local minimum at $$ x = \sqrt{\frac{7}{6}} $$ with value $$ 2\sqrt{42} $$.

The function has one local minimum and one local maximum.
This function has a local maximum at
with value
and a local minimum at
with value
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