#extreme values

Extreme values: the maximum and minimum values. Absolute (global) maximum: the highest y-value possible from x. Absolute (global) minimum: the lowest y-value possible from x. Endpoints: where the function ends. Relative (local) maximum: when the function rises then falls but is not the absolute maximum. Relative (local) minimum: when the function falls then rises but is not the absolute minimum. Critical values: the zeros and undefined points of the derivative. Cusp: when the function has a corner. When extreme values do not exist: when x approaches infinity, including holes, and when there is a cusp.

Derivative test: on a sign diagram when -+ = minimum, +- = maximum

From the image above, determine the a) extreme values, b) endpoints, c) critical values, d) cusps. Question A: We need to find the absolute maximum, absolute minimum, and relative maximas and minimas. Since the absolute maximum is the highest y-value according to x, we look on the graph and see where x = 4. It appears to be the absolute maximum, because there is no cusp and the function does not go off into infinity or negative infinity. Absolute maximum: at 4, the value being f(4) = 5 We do the same procedure with absolute minimum. Absolute minimum: at -3, the value being f(-3) = -2 Relative maximas and miminas are the maximums and minimums on the graph that exclude absolute maximas and miminas. Relative maximum: at 5, the value being f(5) = 2 Relative minimum: at - 4, the value being f(-4) = -1 Question B: Endpoints are when the function has a set domain and does not go off into infinity. This function has two endpoints, and that is at (-4,-1) and (5,2). Question C: The critical values are the zeros of the derivative of the function, or when the derivative of the function is undefined. This question does not give us the equation of the function, so we cannot calculate the critical values. If f(x) = x³ - 3x + 2 f'(x) = 3x² - 3 0 = 3x² - 3 0 = 3(x² - 1) x = ±1 Therefore, our critical values are +-1. Question D: Cusps are like corners on the graph, where it is impossible to draw and calculate the tangent line at that given point. Therefore, we cannot include these as extreme values, but they are still important. Cusps at (-1,2) and (2,6).

Find two positive numbers whose product is 10000 and whose sum is a minimum. We create an equation: S = x + y However, we need just two variables, so we use the information in the question, solve for a variable, and then plug it into the made equation. 10000 = xy y = 10000/x Plugging it into the made equation: S = x + y S = x + 10000/x Differentiate:

Calculate the critical values of the derivative, create a sign diagram and use the derivative test to determine the maximas and minimas.

Creating a sign diagram with indicators:

*Note; We are using dS/dx to determine increase and decrease. We use the derivative test (-+ = minimum, +- = maximum) to conclude 100 is a minima. Since 100 is a minimum value, and the question says the sum is a minimum, we now have S and x. we now can use the other equation to solve for y. y = 10000/x y = 10000/100 y = 100 Therefore, both numbers, x and y, are 100.

If 2700 cm² of material is available to make a box with a square base and open top, find the largest possible volume of the box.

Since we are dealing with volume, we will use the volume equation. V = base · height V = x²h We need to have only one variable though, so we eliminate h by using the given information above, stating the surface area of the material is 2700 cm². SA = base² + 4sides SA = x² + 4xh Solving for h: SA - x² = 4xh h = SA - x²/4x h = 2700 - x²/4x Plugging h into the volume equation: V = x²h V = x²(2700 - x²/4x) Simplify:

Differentiate:

Find the critical values, create a sign diagram with indicators, and use the derivative test to determine the maximas and minimas. Finding the critical values:

Creating sign diagram with indicators: