Question

Find the extreme values of the function on the given interval. \(f(x)=x^{3}-\frac{9}{2} x^{2}-30 x+3\) on [0,6] .

Short answer

The maximum is 3 at \(x=0\) and the minimum is -108 at \(x=6\).

Step-by-step solution

Find the Derivative

First, we need to find the derivative of the function \( f(x) = x^3 - \frac{9}{2}x^2 - 30x + 3 \). Use power rule to differentiate:\[ f'(x) = 3x^2 - 9x - 30 \]

Set Derivative to Zero for Critical Points

To find critical points, set the derivative equal to zero:\[ 3x^2 - 9x - 30 = 0 \]Solve this quadratic equation using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 3 \), \( b = -9 \), and \( c = -30 \).

Solve the Quadratic Equation

Substitute \( a = 3 \), \( b = -9 \), and \( c = -30 \) into the quadratic formula:\[ x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \times 3 \times (-30)}}{2 \times 3} \]Simplify:\[ x = \frac{9 \pm \sqrt{81 + 360}}{6} \]\[ x = \frac{9 \pm \sqrt{441}}{6} \]\[ x = \frac{9 \pm 21}{6} \]The solutions are \( x = 5 \) and \( x = -2 \). Since \( x = -2 \) is not in the interval [0, 6], we only consider \( x = 5 \).

Evaluate at Critical Point and Endpoints

Evaluate the function \( f(x) \) at the critical point and at the endpoints of the interval:\[ f(0) = (0)^3 - \frac{9}{2}(0)^2 - 30(0) + 3 = 3 \]\[ f(5) = (5)^3 - \frac{9}{2}(5)^2 - 30(5) + 3 = -84.5 \]\[ f(6) = (6)^3 - \frac{9}{2}(6)^2 - 30(6) + 3 = -108 \]

Identify Extreme Values

Compare the function values at \( x = 0 \), \( x = 5 \), and \( x = 6 \):- \( f(0) = 3 \)- \( f(5) = -84.5 \)- \( f(6) = -108 \)The highest value is at \( x = 0 \) with \( f(0) = 3 \) (maximum), and the lowest value is at \( x = 6 \) with \( f(6) = -108 \) (minimum).

Key concepts

Critical Points

Critical points are key to understanding the behavior of a function. These points occur where the derivative of the function equals zero or is undefined. To spot critical points, we first calculate the derivative of the function.

• This involves differentiating the given function using simple differentiation rules.
• After finding the derivative, we set it equal to zero and solve for the variable.

Finding these critical points helps us locate potential maximums or minimums of the function, often called extreme values. They are crucial because they indicate where the slope of the tangent to the curve is horizontal, which is where local extreme values may occur. Additionally, these points help in sketching the graph of the function and understanding its increasing or decreasing behavior.

Derivative

A derivative is a fundamental concept in calculus that represents the rate of change of a function's value with respect to a change in its input. It's like answering how the function behaves or changes at any point on its graph.

• The derivative of a function is found using rules like the power rule, product rule, and chain rule, depending on the type of function you're dealing with.
• For example, for the function provided, we apply the power rule to each term individually: \[f'(x) = 3x^2 - 9x - 30\]

The derivative gives us critical points, as we set it equal to zero. This is because we are interested in where the function doesn't increase or decrease momentarily. Simplifying the derivative allows us to better understand the curve's behavior across different intervals.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two and has the form \(ax^2 + bx + c = 0\). Solving these equations is a common task in mathematics to find the roots (or solutions) which can represent points like critical points on a graph.

• In our problem, the equation \(3x^2 - 9x - 30 = 0\) emerges from simplifying the derivative to find critical points.
• We then use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula helps find the values of \(x\) where the function's slope is zero.

These roots are essential for identifying where the potential extreme values are located on the function's graph. Notably, in practical applications, only the roots within the specified interval are considered relevant.