In a problem like, Find the critical point of $$f(x)= \frac{5x}{x-3}$$ What happens to the $5x$ in the numerator? I see that my book has the answer as 3. in order to get 3 I see that I can set $x-3=0$ and add 3 to both sides correct $x-3+3=0+3$ Therefore $x=3$. Is this the correct way to look at this problem?

# [Math] Critical point of a Fraction

calculus

#### Related Solutions

What is the concrete definition of critical point?

Before giving the concrete definition, it's better to have a rough idea about critical points.

Critical pointsare points at which an extremum couldpossiblyoccur.

*Source: Â© CalculusQuestâ„¢*

*Image sources: Solomon Xie's story, http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png*

Therefore, it makes sense to includes three types of points in the domain

- points at which the derivative of $f$
*vanishes*(local min/max, points of inflection) *endpoints*of domain- points at which the derivative of $f$ is
*undefined*(corner, cusp, points of discontinuity)

On a math test, we were instructed to find critical points of the function $f(x) = x\sqrt{30-x^2}$.

The given domain of $f$ is *not* clearly stated in the question. The open interval $(-\sqrt{30},\sqrt{30})$ is just OP's *perception*. Taking account of OP's comment and of the 3rd paragraph of the question body, it *seems* that OP has *mistaken the domain of* $f$, which should actually be the *closed and bounded interval* $[-\sqrt{30},\sqrt{30}]$.

Find the critical points type by type.

- $f'(x) = \sqrt{30 - x^2} - x \, \dfrac{x}{\sqrt{30 - x^2}} = \dfrac{30 - 2x^2}{\sqrt{30 - x^2}}$, so $f'(x) = 0$ iff $x = \pm\sqrt{15}$
- endpoints of the domain of $f$: $x = \pm\sqrt{30}$
- The denominator of $f'(x)$ vanishes iff $x^2 = 30$, i..e the endpoints of the domain of $f$.

Conclusion: The critical points of $f$ are $x = \pm\sqrt{15}, \pm\sqrt{30}$.

# Alternative solution 1

Observe that $f$ is an odd function, since it's a product of an odd function $x \mapsto x$ and an even function $x \mapsto \sqrt{30 - x^2}$. Therefore, it suffices to find the global maximum of $f$. Since $f$ vanishes at the endpoints and the midpoint of the domain, the global extrema are actualy local extrema. The fact that $f$ is odd allows us to concentrates on nonnegative real numbers and apply a.m.-g.m.-inequality
$$\frac{a^2 + b^2}{2} \ge ab$$ with $a = x$ and $b = \sqrt{30 - x^2}$.
\begin{align}
\frac{x^2 + (30 - x^2)}{2} \ge& x \sqrt{30 - x^2} \\
x \sqrt{30 - x^2} \le& 15
\end{align}
Equality holds iff $a = b$.
\begin{align}
\text{i.e. } \quad x &= \sqrt{30 - x^2} \\
x^2 &= 30 - x^2 \\
x &= \pm\sqrt{15}
\end{align}
This elementary solution in algebra-precalculus is useful in contest-math since you *don't* need to use calculus.

# Alternative solution 2

It's *even simpler* to make use of quadratics. Note that in the right half of the domain, everything is *nonnegative*, so squaring $f$ *won't* affect the answer. As a result, consider $$(f(x))^2 = x^2 (30 - x^2) = -(x^2 - 15)^2 + 15^2 \le 15^2.$$ This gives the *same* maximizer $x = \sqrt{15}$ as expected.

The definition of the derivative of a function is the rate of change of a function, say $f(x)$. It defines the gradient of the tangent at a specific point $x$. For a parabola, we visualise that the gradient is always changing, so the derivative of a parabola would be $f'(x) = ax+b$.

If the derivative of a function is a constant value, say, $2$, then this implies that the function's rate of change is $2$ through the whole function, for $x \in \mathbb R$. The only graph where this occurs is a linear graph, which is in the form $f(x) = ax+b$. This also happens to be the derivative of a parabola. Obviously, a line that extends infinitely in either direction does not have a critical point.

## Best Answer

Usually you would define a "critical point" as the zero-points of the first derivative of your function. Since this functions first derivative has no zero-point, the critical point you search for is probably the point where your function is not defined. This means for your example to find the zero-points of the denominator, because it is "not allowed" to divide by 0. Your posted solution does exactly this and hence is the correct way to look at this problem.