Question

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant. $$f(x)=x^{3}-3 a x^{2}+3 a^{2} x-a^{3}$$

Short answer

Answer: The critical point of the function is $$x=a$$.

Step-by-step solution

Find the derivative of the function

Using the power rule, we find the derivative of the given function. $$f'(x) = \frac{d}{dx}(x^3 - 3ax^2 + 3a^2x - a^3)$$ Applying the power rule to each term, we get: $$f'(x) = 3x^2 - 6ax + 3a^2$$

Set the derivative equal to zero and solve for x

Now that we have the derivative, we need to find the values of x where it equals zero. $$3x^2 - 6ax + 3a^2 = 0$$ We can divide the entire equation by 3, which simplifies it to: $$x^2 - 2ax + a^2 = 0$$ This is a quadratic equation in the form of $$ax^2 + bx + c = 0$$, and it can be factored by looking for two numbers that multiply to $$a^2$$ and add up to $$-2a$$. In this case, the two numbers are -a and -a. Therefore, the factored form is: $$(x-a)(x-a) = 0$$

List the critical points

Finally, we can find the critical points by setting each factor equal to zero and solving for x: 1. Case 1: $$(x-a) = 0$$, then $$x = a$$ Since there is only one unique factor, the function has only one critical point. So, the critical point of the function $$f(x) = x^3 - 3ax^2 + 3a^2x - a^3$$ is: $$x=a$$

Key concepts

Understanding Derivatives

A derivative is a core concept in calculus that reflects how a function changes at any point. In simpler terms, it is the slope of the tangent line at a particular point on a graph. The derivative measures the rate of change and helps us locate the critical points of a function.

For the given function, which is a polynomial, we use the power rule to find the derivative. The power rule is straightforward: for any term of the form \(x^n\), the derivative is \(nx^{n-1}\).

• The derivative of \(x^3\) is \(3x^2\).
• The derivative of \(-3ax^2\) is \(-6ax\).
• The derivative of \(3a^2x\) is \(3a^2\).
• The derivative of a constant \(-a^3\) is \(0\).

Putting these together gives the derivative \(f'(x) = 3x^2 - 6ax + 3a^2\).

This expression allows us to investigate how the function behaves at various points along the x-axis.

Quadratic Equation Insights

Once we've derived the expression for the derivative, we need to find its critical points. This usually involves solving an equation, often a quadratic equation. Quadratic equations are polynomial equations of degree 2, which are expressed in the form \(ax^2 + bx + c = 0\).

In our case, the equation becomes \(3x^2 - 6ax + 3a^2 = 0\). To simplify, we divide the entire equation by 3, resulting in \(x^2 - 2ax + a^2 = 0\).

• Quadratic equations can be solved by methods like factorization, completing the square, or using the quadratic formula.
• In this instance, factorization is the most efficient approach due to its simplicity.

Understanding these equations is vital for finding where the slope of the original function is zero, leading to critical points.

The Role of Factorization

Factorization involves breaking down an equation into simpler parts or expressions that can be multiplied together to give the original equation. For quadratic equations, this often means finding two binomials that multiply to form the original expression.

In the given equation \(x^2 - 2ax + a^2 = 0\), we factor it by identifying numbers that multiply to \(a^2\) and add to \(-2a\). These numbers are \(-a\) and \(-a\), resulting in the factorized form \((x-a)(x-a) = 0\).

• This indicates the critical point where the derivative and, hence, the slope of the tangent is zero.
• By setting each factor equal to zero, we solve for \(x = a\).

Factorization eases the process of finding solutions to quadratic equations and is instrumental in locating critical points of a function.