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!!!
$$\begin{align}
f'(x) &= \frac{(8x^2+x-3)^{-1/2}}{2}\frac{d}{dx}(8x^2+x-3)\\
&= \frac{(8x^2+x-3)^{-1/2}}{2}(16x+1)\\
&= \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}
\end{align}$$
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 $8(\frac{-1}{16})^2+\frac{-1}{16}-3<0$, so taking its square root will give us an imaginary number. Thus the answer is DNE
 A: 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 $$\frac{-1-\sqrt{97}}{16} < \frac{-1}{16}$$
we see that $x=-1/16$ is not in the domain of $f(x)$. So, $f(x)$ doesn't have any horizontal tangent lines.
A: Shortcut:
$x\mapsto \sqrt x$ has non-zero derivative everywhere, hence $f(x)=\sqrt{8x^2+x-3}$ having a horizontal tangent is the same as $g(x)=8x^2+x-3$ having a horizontal tangent for the same $x$, and $g(x) \geqslant 0$ for that $x$ so that we can take the square root.
$g'(x)=16x+1$ which is zero for $x_0=-1/16$.
$g(x_0) = \tfrac8{16^2}-\tfrac1{16}-3 = \tfrac1{32}-\tfrac1{16}-3 = -\tfrac1{32}-3  < 0$ hence no horizontal tangent.
The first is intuitive, but you also see it from 
$$\big(u(v(x))\big)' = v'(x)\cdot u'(v(x))$$
If $u'$ is non-zero everywhere, then the only way that the product can be zero is when $v'$ is zero.
A: We know the answer by calculus.  Here an alternative method is explored, based on analytic geometry.
Recognizing the graph as (part of) a conic curve, we square both sides of the equation and bring the result into standard form:
$8x^2-y^2+x-3=0$
The quadratic terms factor as $(2\sqrt2x+y)(2\sqrt2x-y)$ so the conic is, of course, a hyperbola, and the full equation of the hyperbola will have the product form
$(2\sqrt2x+y+a)(2\sqrt2x-y+b)=c$
Expanding the left side and matching like terms to the standard form equation rendered above leads to
$(2\sqrt2)(a+b)=1$ matching linear terms in $x$
$b-a=0$ matching linear terms in $y$
$ab-c=-3$ matching constant terms
The two linear equations give $a=b=1/(4\sqrt2)$, then the last equation gives $c=97/32>0$.  Thus
$(2\sqrt2x+y+(1/(4\sqrt2)))(2\sqrt2x-y+(1/(4\sqrt2)))=97/32$
We now draw the asymptote given by each factor on the left being set to zero; then since their product is actually positive the hyperbola must lie in the regions where the factors have identical signs.  This corresponds to the shaded regions in the sketch below (not drawn to scale):

We see that the horizontal line through the center cuts through the allowed quadrants and thus through the actual hyperbola, and then any other horizontal line similarly cuts through.  There cannot be any horizontal tangents because the hyperbola ends up in the wrong pair of "quadrants" defined by its asymptotes.
To have the hyperbola in the right quadrants and get a horizontal tangent we would have needed $c$ to be negative.  This would have been equivalent to identifying a nonzero real value of $y$ at the zero derivative point rendered by calculus.  Instead of that happy ending, the actual positive value of $c$ corresponds to an imaginary value of $y$ instead and thus no (real) horizontal tangent.
A: 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., $y\in\mathbb{R}$.
