Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line Section 2.5 #14

Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line.

Okay, so having a horizontal tangent line at a point on the graph means that the slope of that tangent line is zero. The derivative of a function is another function that tells us the slope of the tangent line at any given point on the graph of the original function. 

Thus, to find where the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line, we need to take the derivative, set it equal to zero, and solve for $x$. This will give us the $x$-coordinate of where the graph of $f(x)$ has a horizontal tangent line. To find the corresponding $y$ value, we plug the $x$ value that we found into the original equation. 

In this problem, when we plug the $x$ value we find into the original equation, we get an imaginary number, which means that no point on the graph of $f(x)$ has a horizontal tangent line, and thus our answer is DNE, does not exist. Let's go through the motions!!!

$f(x) = \sqrt{8x^2+x-3}=(8x^2+x-3)^{1/2}$

$f'(x) = \frac{d}{dx}(8x^2+x-3)^{1/2}$

Time do the chain rule!!!

$$\begin{align}
f'(x) &amp;= \frac{(8x^2+x-3)^{-1/2}}{2}\frac{d}{dx}(8x^2+x-3)\\
&amp;= \frac{(8x^2+x-3)^{-1/2}}{2}(16x+1)\\
&amp;= \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}
\end{align}$$

Alright, we have our derivative. We want to find horizontal tangent lines, so we set this equal to zero and solve for $x$

$$0 = \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}$$

multiplying both sides of the equation by $2(8x^2+x-3)^{1/2}$ we get

$0 = (16x+1)$

And thus $x = \frac{-1}{16}$

Now, we plug this value into the original equation to get the corresponding $y$ value, because remember, we are looking for a point on the graph where the horizontal line is tangent, so our answer will be in $(x,y)$ format, is it exists, (which in this case, it won't)..

$f(\frac{-1}{16}) = \sqrt{8(\frac{-1}{16})^2+\frac{-1}{16}-3}$

But $8(\frac{-1}{16})^2+\frac{-1}{16}-3&lt;0$, so taking its square root will give us an imaginary number. Thus the answer is DNE
Given

$$f(x) = \sqrt{8x^2+x-3}$$

the horizontal tangent lines of $f(x)$ occur all at points in the domain of $f(x)$ where $f'(x)=0$. You have correctly found the derivative

$$f'(x)=\frac{16x+1}{2\sqrt{8x^2+x-3}}$$

which is a rational function. The zeroes of a rational function are where the numerator is zero. Solving $16x+1=0$, we find that $x=-1/16.$

In order to have a horizontal tangent line at $x=-1/16$, $f(x)$ must be defined at $x=-1/16$.
The domain of $f(x)=\sqrt{8x^2+x-3}$ is where $8x^2+x-3\ge0$. Solving this inequality, we find the domain of $f(x)$ as

$$x\ge \frac{1}{16}\left(\sqrt{97}-1\right)$$
$$x\le \frac{1}{16}\left(-1-\sqrt{97}\right)$$

therefore since $$\frac{-1-\sqrt{97}}{16} &lt; \frac{-1}{16}$$

we see that $x=-1/16$ is not in the domain of $f(x)$. So, $f(x)$ doesn't have any horizontal tangent lines.
First off, I assume you are working in $\mathbb{R}$, yes?  What is the domain of your original function?  Since $$f\left(x\right)=\sqrt{8x^{2}+x-3}\text{, the domain is implied: }8x^{2}+x-3\ge 0.$$Solve this to find the domain.  Then, when you find the $x$-value of the point where the tangent is $0$, see if that point is in the domain and if it is, you should end up with a real $y$-value, i.e., $y\in\mathbb{R}$.