For what value of 'x' will the function y = x2 - 4x have the maximum or minimum value?
2
To find the value of 'x' for which the function \(y = x^2 - 4x\) has its maximum or minimum value, we apply the principles of differential calculus. For a function to attain a local maximum or minimum, its first derivative with respect to 'x' must be equal to zero. These points are commonly referred to as critical points.
Let's consider the given function:
\[y = x^2 - 4x\]
Derivative Calculation for Function Optimization
Step 1: Compute the First Derivative
We need to differentiate the function \(y\) with respect to \(x\), denoted as \(\frac{dy}{dx}\). We use the power rule for differentiation, which states that if \(f(x) = ax^n\), then \(f'(x) = nax^{n-1}\).
- For the term \(x^2\): Applying the power rule (\(n=2, a=1\)), its derivative is \(2 \cdot x^{2-1} = 2x\).
- For the term \(-4x\): Applying the power rule (\(n=1, a=-4\)), its derivative is \(1 \cdot (-4) \cdot x^{1-1} = -4x^0 = -4\).
Combining these, the first derivative of the function is:
\[\frac{dy}{dx} = 2x - 4\]
Step 2: Set the First Derivative to Zero
To locate the critical points where a maximum or minimum could occur, we set the first derivative equal to zero and solve for \(x\):
\[2x - 4 = 0\]
Add 4 to both sides of the equation:
\[2x = 4\]
Divide both sides by 2:
\[x = \frac{4}{2}\] \[x = 2\]
This value, \(x = 2\), is the critical point for the function \(y = x^2 - 4x\).
Extrema Determination Using Second Derivative Test
Step 3: Calculate the Second Derivative
To ascertain whether the critical point \(x=2\) corresponds to a maximum or minimum value, we employ the second derivative test. This involves differentiating the first derivative, \(\frac{dy}{dx}\), once more with respect to \(x\).
Our first derivative is \(\frac{dy}{dx} = 2x - 4\).
- Differentiating the term \(2x\): Its derivative is \(2\).
- Differentiating the term \(-4\) (which is a constant): Its derivative is \(0\).
Thus, the second derivative, \(\frac{d^2y}{dx^2}\), is:
\[\frac{d^2y}{dx^2} = 2\]
Step 4: Apply the Second Derivative Test
Now, we examine the sign of the second derivative at the critical point \(x = 2\).
In this particular case, \(\frac{d^2y}{dx^2} = 2\), which is a positive constant value. According to the second derivative test, if the second derivative is positive (\(\frac{d^2y}{dx^2} > 0\)) at a critical point, the function has a local minimum at that point. If it were negative, it would indicate a local maximum.
| Condition for \( \frac{d^2y}{dx^2} \) | Type of Extrema Indicated |
|---|---|
| \( \frac{d^2y}{dx^2} > 0 \) | Local Minimum |
| \( \frac{d^2y}{dx^2} < 0 \) | Local Maximum |
| \( \frac{d^2y}{dx^2} = 0 \) | Inconclusive (Additional analysis required) |
Therefore, for the function \(y = x^2 - 4x\), the maximum or minimum value occurs at \(x = 2\). Specifically, since the second derivative is positive, this value of \(x\) corresponds to a minimum value of the function.
Important Questions from Maxima & Minima
Which of the following statements is false about convex minimization problem?
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As \(\rm x\) varies from \(\rm −1\ to \ +3\), which one of the following describes the behaviour of the function \(\rm f(x) = x^3 – 3x^2 + 1\)?
The function f(x) = 8 loge x - x2 + 3 attains its global minimum over the interval [1, e] at x = ________.
(Here loge x is the natural logarithm of x and e2 = 7.39 )
