Find the points on the curve y=2 x^3+3 x^2-12 x+1, where the tangent is horizontal.

Name: Problem 16, Chapter 8: Differential Calculus Booster with Problems and Solutions for JEE Main and Advanced
Uploaded: 2020-06-04T21:28:27-08:00
Duration: 6 min 56 s
Channel: Kira Schwander
Description: Find the points on the curve y=2 x^3+3 x^2-12 x+1, where the tangent is horizontal.

Find the points on the curve y=2 x^3+3 x^2-12 x+1, where the...

Key Concepts

Critical Points

Critical points are points on a function where the derivative is zero or undefined. They serve as potential locations for local extrema and other significant changes in the graph's behavior, making them important in the study of functions.

Polynomial Function

A polynomial function is an algebraic expression composed of variables and coefficients, involving only non-negative integer exponents. Differentiating polynomials is typically straightforward due to the applicability of the power rule, which simplifies the process of finding the rate of change at various points.

Horizontal Tangent Line

A horizontal tangent line occurs at points on a curve where the derivative, which represents the slope of the tangent, is equal to zero. These points are crucial for identifying local maximums, local minimums, or points of inflection on the graph of a function.

Derivative

The derivative is a fundamental concept in calculus that measures the instantaneous rate of change of a function. It is calculated using differentiation rules, such as the power rule for polynomial functions, and is essential for analyzing the behavior of functions.

Transcript

00:01 Hello.

00:04 I'm going to be working with this curve y equals two x cubed plus three x squared, minus 12 x plus one.

00:12 And we're going to be looking for at which points is the tangent of this line going to be equal to zero? so we're going to start off by going over to dismas and looking at this curve.

00:24 I already put the equation for the curve into dismas.

00:27 So here is the graph of this line.

00:30 So we would expect the tangent to be horizontal at the maximum and minimum on.

00:37 This is because the tangent is the slope and the slope starts steep down here as it gets closer to the maximum.

00:45 It kind of levels out before tipping in the other direction to start going down, and then at the minimum, it levels out to zero and then tips back upward.

00:57 So we're going to expect the tension.

00:59 To be zero at these two points will keep that in mind while we're going through this problem.

01:06 So we're going to start off by finding the first derivative of this equation on.

01:11 That's because the first derivative of a slope of a line is the slope of that line.

01:17 So we're going to do why prime equals for the first term.

01:21 We're going to multiply the constant of two by the exponents of three.

01:25 So we're going to get six x and then the exponents of three subtracted by one is going to be too.

01:34 So that term is going to be six x squared.

01:37 Similarly, for the next 13 times two is six x two minus one is one minus.

01:45 Um, there's, ah, an implied one here.

01:48 So 12 negative, 12 times one is negative, 12 x one minus 10 and then, since there's an implied x to the zero over here, one time zero is gonna be zero.

02:04 So that constants going to cancel out.

02:07 So the tangent is going to be horizontal when the slope of the line is equal to zero.

02:14 So using that, we're going to set this the derivative equal to zero, and we're just going to use this same equation minus 12th.

02:25 We're going to simplify that by factoring out of six because we're trying to, um, we're trying to isolate those excess, and we want it to be a simple as possible zero divided by six is still zero, so we have zero equals x squared plus x minus two.

02:45 So this equation is easily factory herbal because we know that the factors of negative to our positive and negative one and two so we need to find factors that are going to multiply to be negative, too, and add to be positive one for the middle term.

03:05 So we see that if zero equals x minus, one times x plus two negative one times positive two equals negative two and negative one plus two is equal to positive one, so this is going to factor out properly.

03:25 So now that we have this either one of those terms, either x minus one or x plus two is going to need to be equal to zero or both...

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