Question

Find the critical points of the function \(f(x)=\sinh ^{2} x \cosh x\).

Short answer

Answer: The critical point of the function is \(x = 0\).

Step-by-step solution

Find the derivative of the function

Start by finding the derivative of the function \(f(x) = \sinh^2 x \cosh x\). Firstly, we will need to use the product rule, which states that \((uv)' = u'v + uv'\). Let \(u = \sinh^2 x\) and \(v = \cosh x\), then \(u' = 2\sinh x \cosh x\) and \(v' = \sinh x\). We get: $$ f'(x) = (u'v + uv') = (2\sinh x \cosh x)(\cosh x) + (\sinh^2 x)(\sinh x). $$ Now, simplify the expression: $$ f'(x) = 2\sinh x \cosh^2 x + \sinh^3 x. $$

Set the derivative equal to zero

Now that we have the derivative, we can find the critical points by setting the derivative equal to zero and solving for x. $$ 0 = 2\sinh x \cosh^2 x + \sinh^3 x. $$

Factor out the common term

Observe that \(\sinh x\) is the common factor of both terms. Factor \(\sinh x\) out of the expression: $$ 0 = \sinh x (2 \cosh^2 x + \sinh^2 x). $$

Solve for x

Now we have two cases to consider: 1. \(\sinh x = 0 \Rightarrow x = 0\). 2. \(2\cosh^2 x + \sinh^2 x = 0\). Since \(\cosh^2 x \ge 1\) and \(\sinh^2 x \ge 0\) for all real x, it is not possible for \(2\cosh^2 x + \sinh^2 x = 0\). Therefore, there is only one critical point, which is \(x = 0\).

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, but they relate to hyperbolas in the same way trigonometric functions relate to circles. The most common hyperbolic functions used in calculus are the hyperbolic sine, \(\sinh x\), and the hyperbolic cosine, \(\cosh x\). These functions are defined as:\

\
• \(\sinh x = \frac{e^x - e^{-x}}{2}\)
\
• \(\cosh x = \frac{e^x + e^{-x}}{2}\)
\

\
\They have a rich set of properties similar to trigonometric functions, such as identities like \(\cosh^2 x - \sinh^2 x = 1\). When dealing with derivatives and integrals, their properties allow for a smooth transition of concepts from trigonometry. In the exercise above, these hyperbolic functions play a key role as we find critical points involving both \(\sinh x\) and \(\cosh x\). Understanding these functions provides valuable tools for solving a wide range of algebraic problems involving exponential functions.

Product Rule in Calculus

The product rule is an essential tool in calculus for finding the derivative of the product of two functions. Suppose you have two differentiable functions, \(u(x)\) and \(v(x)\). The product rule states that the derivative of their product is given by:

• \((uv)' = u'v + uv'\)

In simpler terms, to differentiate the product of two functions, differentiate each function separately and multiply by the other function, then sum those products. This rule is crucial when dealing with expressions where one part of the product contains a complicated function that needs to be derived. In the exercise about the function \(f(x) = \sinh^2 x \cosh x\), the product rule helps in breaking down the derivative into manageable parts and ensures an accurate solution for determining critical points. Using the product rule correctly allows for a clearer understanding of how each part of the function contributes to the overall derivative.

Derivative of a Function

A derivative represents how a function changes as its input changes. It is a fundamental concept in calculus that helps in understanding rates of change and slopes of curves. More formally, if \(f(x)\) is a function, the derivative \(f'(x)\) gives the rate at which \(f(x)\) is changing at any point \(x\). The process of finding a derivative is called differentiation.

• For basic functions like polynomials, differentiation is straightforward using rules like the power rule.
• For more complex functions, rules such as the product rule and chain rule come into play.

In the given exercise, finding the derivative \(f'(x) = 2\sinh x \cosh^2 x + \sinh^3 x\) was the first step in identifying critical points. This derivative gives insight into where the function \(f(x)\) has slopes of zero, indicating potential maximum, minimum, or saddle points. Understanding how to compute and simplify derivatives is vital when analyzing functions and solving real-world problems.