How do you use the first and second derivatives to sketch #y=(x^3)-(6x^2)+5x+12#?

Question

Christopher Allers · Accepted Answer

The first derivative allows you to determine the location of the critical points of the function. The second allows you to determine the nature of these points.

The first derivative #(dy)/(dx)# describes how y changes with x, ie a gradient. The second derivative #(d^2y)/(dx^2)# effectively describes how the gradient changes with x, ie the curvature. At a turning point, the gradient will be zero, so the turning points can be located by setting the first derivative equal to zero. #y' = 3x^2 - 12x + 5 = 0# Using the quadratic formula: #x = (12 +- sqrt(144 - 60))/6# #x = 3.528 or x = 0.472# This gives the location of the turning points, we now find the second derivative to determine their natures. #y'' = 6x - 12# We can see that for x = 3.528 that y'' > 0, x = 0.472 that y'' < 0. For positive curvature, it will be curving upwards (think a positive, happy smiley face :)) so the point there will be in a trough, ie a minimum. The point with negative curvature will be curving downwards so will be a maximum. With this, plug in the x values to find the values of the function at these critical points, mark them taking note of their natures and then, taking into account other useful features like intercepts and asymptotes, plot the graph.

Ezekiel Alexander · Answer

undefined To sketch the curve $ y = x^3 - 6x^2 + 5x + 12 $ using the first and second derivatives:

1. Find the first derivative $ y' $:
$$ y' = 3x^2 - 12x + 5 $$

2. Set $ y' = 0 $ to find critical points:
$$ 3x^2 - 12x + 5 = 0 $$
Solve for $ x $ to find the x-values of the critical points.

3. Use the second derivative test to determine the nature of the critical points:
$$ y'' = 6x - 12 $$
- If $ y'' > 0 $, the function is concave up at that point (local minimum).
- If $ y'' < 0 $, the function is concave down at that point (local maximum).
- If $ y'' = 0 $, the test is inconclusive.

4. Sketch the curve:
- Plot the critical points and the points where the first derivative is zero.
- Use the information from the second derivative test to determine the concavity between the critical points.
- Sketch the curve passing through these points, considering the concavity and the behavior of the function as $ x $ approaches infinity or negative infinity.