91Ó°ÊÓ

Determining critical points is an essential step in optimization problems in calculus. It's where the function's slope is zero, indicating potential maximum or minimum values. First, you need the first derivative of the function.In this exercise, we have the function: \( B(x) = 305x^2 - 1830x^3 \).Compute the first derivative using the power rule:\( B'(x) = 610x - 5490x^2 \).To find critical points, set the derivative equal to zero:\( 610x - 5490x^2 = 0 \).Solve the equation: factor out common terms, which gives us two critical points, \( x = 0 \) and \( x = \frac{1}{9} \). Understanding where these critical points occur allows us to move on to the next steps, determining if these points indicate a maximum or minimum value.

The first derivative of a function provides us with the slope of the tangent line to the curve at any point. It's crucial in finding where the function's rate of change is zero, indicating critical points. For this exercise, the first derivative is: \( B'(x) = 610x - 5490x^2 \).By setting \( B'(x) = 0 \), we can solve for the values of \( x \), which are our critical points. The solutions are \( x = 0 \) and \( x = \frac{1}{9} \). These critical points will be used for further analysis using the second derivative to determine whether they correspond to a maximum or minimum.

The second derivative test helps us determine the concavity of the function at the critical points, which tells us if they are maxima, minima, or points of inflection. For the function \( B(x) = 305x^2 - 1830x^3 \), we first find the second derivative:\( B''(x) = 610 - 10980x \).Next, substitute the critical points into the second derivative:

• At \( x = 0 \), \( B''(0) = 610 \)
• At \( x = \frac{1}{9} \), \( B''\bigg(\frac{1}{9}\bigg) = 610 - 10980\bigg(\frac{1}{9}\bigg) = -610 \)

If \( B''(x) \) is positive at a critical point, the function has a local minimum at that point. If \( B''(x) \) is negative, the function has a local maximum. Here, \( x = 0 \) is a local minimum, and \( x = \frac{1}{9} \) indicates a local maximum.

After confirming that \( x = \frac{1}{9} \) is a local maximum using the second derivative test, we need to find the actual value of the maximum blood pressure. Substitute \( x = \frac{1}{9} \) back into the original function:\( B\big(\frac{1}{9}\big) = 305\bigg(\frac{1}{9}\bigg)^2 - 1830\bigg(\frac{1}{9}\bigg)^3 \).Simplify the expression step-by-step:

• \( B\big(\frac{1}{9}\big) = 305 \times \frac{1}{81} - 1830 \times \frac{1}{729} \)
• \( B\big(\frac{1}{9}\big) = \frac{305}{81} - \frac{1830}{729} \)
• \( B\big(\frac{1}{9}\big) = \frac{305}{81} - \frac{610}{243} \)

After simplification, you'll obtain the precise value of the maximum blood pressure at \( x = \frac{1}{9} \). This step ensures we have not only identified the critical point but have accurately calculated the peak blood pressure.