Question

Find the hottest and coldest points on a bar of length 5 if \(T=4 x-x^{2},\) where \(x\) is the distance measured from the left end.

Short answer

The hottest point is at \( x = 2 \), and the coldest point is at \( x = 5 \).

Step-by-step solution

- Identify the function and domain

The temperature function provided is given by the equation \[ T(x)=4x-x^2 \] and the bar length is 5. Therefore, the domain is \[ 0 \leq x \leq 5 \].

- Find the derivative

To find the critical points, take the derivative of the temperature function with respect to \( x \): \[ T'(x) = 4 - 2x. \]

- Set the derivative to zero

Set the derivative equal to zero to find the critical points: \[ 4 - 2x = 0. \] Solving for \( x \): \[ x = 2. \]

- Evaluate the temperature at the critical point and endpoints

Calculate the temperature at \( x = 2 \) and at the endpoints \( x = 0 \) and \( x = 5 \): For \( x = 2 \): \[ T(2) = 4(2) - (2)^2 = 8 - 4 = 4. \] For \( x = 0 \): \[ T(0) = 4(0) - (0)^2 = 0. \] For \( x = 5 \): \[ T(5) = 4(5) - (5)^2 = 20 - 25 = -5. \]

- Identify the hottest and coldest points

Compare the temperature values. The hottest point is at \( x = 2 \) with a temperature of 4. The coldest point is at \( x = 5 \) with a temperature of \(-5\).

Key concepts

Derivative

To find the hottest and coldest points on the bar, we need to use the concept of derivatives in calculus. Derivatives help us analyze the rate at which a function is changing. The temperature function is given by \[ T(x) = 4x - x^2 \]. Taking the derivative of this function with respect to x helps us find where the rate of change is zero, which are known as critical points.

The derivative of the function is found as follows:
\[ T'(x) = \frac{d}{dx}(4x - x^2) = 4 - 2x \]

Setting this derivative equal to zero helps us find the x-values where the slope of the tangent to the curve is zero (i.e., the 'flat' points).
\[ 4 - 2x = 0 \]
Solving this equation:
\[ x = 2 \]
This means at x=2, the temperature function has a critical point. These critical points are essential as they help us identify our maximum and minimum values within the interval.

Domain

The domain of a function determines the set of possible input values (x-values) for which the function is defined. In this exercise, we consider the bar of length 5 units, so the domain is from 0 to 5.

The temperature function \[ T(x) = 4x - x^2 \] is therefore defined for:
\[ 0 \leq x \leq 5 \]

The domain is crucial as it sets the limits within which we analyze the function. Any value of x outside this interval would not be physically meaningful in the context of measuring the bar's temperature. It’s important to evaluate the function at the endpoints of the domain as well, since these endpoints might hold the maximum or minimum values in some cases.
Within this domain, we are working to find where the temperatures are the highest and lowest.

Endpoints

After finding the critical points, we must also consider the function's values at the domain’s endpoints. For this problem, the endpoints are x=0 and x=5. Let’s evaluate the temperature function at these points:

• At x=0: \[ T(0) = 4(0) - (0)^2 = 0 \]
• At x=5: \[ T(5) = 4(5) - (5)^2 = 20 - 25 = -5 \]

Now we have three points to compare:

• x=0: Temperature is 0
• x=2: Temperature is 4 (this is from our critical point calculation)
• x=5: Temperature is -5

By comparing these temperatures, we know the hottest point on the bar is at x=2 with a temperature of 4, and the coldest point is at x=5 with a temperature of -5.
Critical points and endpoints both play a crucial role in identifying the maximum or minimum values of a function over a given interval.