Find the parameter \(m\) for the function \(y = f(x)\) to reach an extremum at \(x_0\)

Post by | Date September 17, 2023 | 0 comment | 191 lượt xem | Facebook:

We know that if the function \(y = f(x)\) has a derivative at \(x_0\) and reaches an extremum at \(x_0\), then \(f'(x_0) = 0\). From this observation, we can outline the steps to solve the problem of finding the parameter \(m\) for the function \(y = f(x)\) to reach an extremum at \(x_0\).

  1. Step 1. Find the domain.
  2. Step 2. Find \(y’\).
  3. Step 3. Solve the equation \(y'(x_0) = 0\) to find the parameter \(m\).
  4. Step 4. Recheck the obtained \(m\) value using the variation table or \(y”\).

Example: Find \(m\) for the function \(y = x^3 – 2mx^2 + mx + 1\) to reach a minimum at \(x = 1\).

Solution:

To have \(x = 1\) as the minimum point of the function, \(y'(1) = 0 \Leftrightarrow m = 1\).

Testing with \(m = 1\), we have \(y’ = 3x^2 – 4x + 1\).

\(y’ = 0 \Leftrightarrow 3x^2 – 4x + 1 = 0 \Leftrightarrow \left\{ \begin{array}{l}x = 1\\x = \frac{1}{3}\end{array} \right..\)

Variation table:

Observing the variation table, we see that \(m = 1\) satisfies the problem’s requirements.

Note: For a cubic polynomial, we can use the following method:

The function \(f(x)\) reaches a maximum at \(x_0\) if and only if \(\left\{ \begin{array}{l}f'(x_0) = 0\\f”(x_0) < 0\end{array} \right.\). The function \(f(x)\) reaches a minimum at \(x_0\) if and only if \(\left\{ \begin{array}{l}f'(x_0) = 0\\f''(x_0) > 0\end{array} \right.\).

Example: Find the real value of the parameter \(m\) for the function \(y = \frac{1}{3}x^3 – mx^2 + (m^2 – 4)x + 3\) to reach a maximum at \(x = 3\).

Solution:

We have \(y’ = x^2 – 2mx + (m^2 – 4)\) and \(y” = 2x – 2m\).

The function \(y = \frac{1}{3}x^3 – mx^2 + (m^2 – 4)x + 3\) reaches a maximum at \(x = 3\) if and only if: \(\left\{ \begin{array}{l}y'(3) = 0\\y”(3) < 0\end{array} \right.\)

\(\Leftrightarrow \left\{ \begin{array}{l}9 – 6m + m^2 – 4 = 0\\6 – 2m < 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m^2 - 6m + 5 = 0\\m > 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = 1\\m = 5\end{array} \right. \Leftrightarrow m = 5\).

So, \(m = 5\) is the value we are looking for.

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