91Ó°ÊÓ

Derivatives are a fundamental tool in calculus. They help us understand how a function changes at any given point. To find the derivative of a function, we apply differentiation rules. The function from the exercise is given by \( f(x) = x^2 + \frac{16}{x} \). To differentiate \( f(x) \), we use the power rule for \( x^2 \) and the quotient rule for \( \frac{16}{x} \).
The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Thus, the derivative of \( x^2 \) is \( 2x \).
For \( \frac{16}{x} \), rewrite it as \( 16x^{-1} \) and apply the power rule to get \( -16x^{-2} \). With these calculations, the derivative \( f'(x) \) becomes:

• \( f'(x) = 2x - \frac{16}{x^2} \)

Derivatives set the stage for identifying critical points and further evaluations.

Critical points are where the function could have a maximum, minimum, or inflection point. These points occur where the derivative is zero or undefined. By finding critical points, we can investigate these potential extremum points.
To locate them, set \( f'(x) = 0 \) and solve:

• \( 2x - \frac{16}{x^2} = 0 \)
• Multiply by \( x^2 \) to clear the fraction, resulting in \( 2x^3 - 16 = 0 \).
• Solve for \( x \), yielding \( x^3 = 8 \), so \( x = 2 \).

This gives us the critical point at \( x = 2 \). At this critical point, further evaluation helps determine if it's a maximum or minimum when combined with endpoint evaluations.

Endpoints of an interval are also important when determining maximum and minimum values of a function. The endpoints in our exercise are \( x = 1 \) and \( x = 4 \). We evaluate the function at these points in addition to the critical points.
Substitute the endpoint values into the function:

• At \( x = 1 \), \( f(1) = 1^2 + \frac{16}{1} = 17 \).
• At \( x = 4 \), \( f(4) = 4^2 + \frac{16}{4} = 20 \).

Both endpoints are evaluated alongside the critical point evaluation \( f(2) = 12 \). Reviewing these allows us to ascertain the overall maxima and minima on the interval [1,4].

Determining maximum and minimum values involves comparing the results from both critical points and endpoint evaluations. The results from substituting these values into the function are:

• \( f(1) = 17 \)
• \( f(2) = 12 \)
• \( f(4) = 20 \)

The maximum value is the highest result, which is \( f(4) = 20 \). The minimum value is the lowest, which is \( f(2) = 12 \). By capturing and comparing these, we've fully utilized calculus optimization to find our extremum points effectively on the interval [1,4].