Cracking The Ap Calculus ab Exam 2018

yourself. Once you have the equation, you find its derivative and set it equal to zero. The values you get ...

This tells us that 11 is a critical point of the equation. Now we need to figure out if this is ...

After we cut out the squares of side x and fold up the sides, the dimensions of the box will be width: 18 − ...

= 24x − 168 At the end of the day, no matter how complex the math might get, if a problem is based on ...

At x = , y = ≈ −0.385 Now it’s time to check the endpoints of the interval. At x = −3, y = ...

The derivative is not defined at x = ±4. Setting the derivative equal to zero, we get 2(16 − x) = 2x^2 32 ...

If we call the length of the field y and the width of the field x, the formula for the area of the field ...

and bottom and a 2-inch margin on each side. Find the overall dimensions that will minimize the total ...

This is positive when x is positive, so the minimum area occurs when x = . Thus, the ...

This is positive when x is positive, so the minimum surface area occurs when x = 8. The dimensions ...

= 2x − 7 = 0 x = Solving for y, we get y = . Finally, because = 2, the point is the min ...

The range of a projectile is , where v 0 is its initial velocity, g is the acceleration due to gravit ...

it curves downward. Step 4: Test End Behavior Look at what the general shape of the graph will be, based ...

(3)The y-coordinates of each critical point are found by plugging the x-value into the original equation. A ...

y = (0)^3 − 12(0) = 0 The curve has a y-intercept at (0, 0). There are no asymptotes, because t ...

Example 2: Sketch the graph of y = x^4 + 2x^3 − 2x^2 + 1. Step 1: First, let’s find the x-int ...

Set the derivative equal to zero. 4 x^3 + 6x^2 − 4x = 0 2 x(2x^2 + 3x − 2) = 0 2 x(2x − 1)(x + ...

or a minimum. 12(0)^2 + 12(0) − 4 = −4 This is negative, so the curve has a maximum at (0, 1); the ...

Example 3: Sketch the graph of y = 2 − x. Step 1: Find the x-intercepts. 2 − x = 0 x = 2 x = ± ...

Again, there’s no x-value where this is zero. In fact, the second derivative is positive at all values of ...