Find the point(s) at which the tangent to the curve is horizontal. a. y=(2 x^2)/(x-4) b. y=(x^2-

Find the point(s) at which the tangent to the curve is...

Key Concepts

Critical Points

Critical points are values in the domain of a function where its derivative is zero or undefined. In the context of finding horizontal tangents, identifying critical points by setting the derivative equal to zero is key, as these points are candidates for local extrema or points of inflection that often correspond to horizontal tangents.

Horizontal Tangent

A horizontal tangent is a line that touches a curve at a point where the slope is zero. This indicates that at that specific point, the rate of change of the function's value with respect to its variable is zero, signaling a potential local maximum, minimum, or a point of inflection.

Derivative

The derivative represents the instantaneous rate of change of a function and is foundational in calculus. It provides information about the slope of a function's graph at any given point, helping identify where the function is increasing, decreasing, or has horizontal tangents when the derivative equals zero.

Quotient Rule

The quotient rule is a technique used in differentiation for functions that are ratios of two differentiable functions. It is essential for calculating the derivative of rational functions, ensuring accurate determination of slopes and aiding in the process of finding horizontal tangents.

Transcript

00:01 Looking at this question, i see that it says that the tangent of the curve is horizontal.

00:06 That means that the slope is zero for a horizontal line.

00:13 So we've got to calculate the slopes, which are the derivatives, of each of these.

00:28 Derivative is going to be denominator times derivative of the numerator, which is going to be 4x, minus numerator, derivative of the denominator, which is one over denominator squared.

00:54 Okay, simplifying this, distributing the 4x, it gives me 4x squared, and then subtracting 2x squared, this just gives me 2x squared, and then minus 16x.

01:14 Okay, if we want the slope to be zero, then the numerator has to be zero.

01:37 And so i can factor this out.

01:42 I'm going to factor out of 2x.

01:50 Either x equals zero, using the zero product property, or x minus 8 equals 0, so x would be 8.

02:00 Now, i have to put these numbers back into the equation.

02:06 So y of 0 is 2 times 0 squared over 0 minus 4, which would be negative 1 half.

02:18 Y of 8 would be 2 times 8 squared.

02:24 8 squared is 64 times 2.

02:27 Well, i'm just going to leave it there.

02:29 2 times 64 would be 128, but i'm just going to leave it there.

02:32 And 8 minus 4 is 4.

02:38 Okay.

02:38 Okay, taking out a 2, and then 64 divided by 2 is 32.

02:47 So the answers are 0, negative, 1, 1, and 8, 32.

03:00 Let's look at b.

03:09 X squared plus x minus 2.

03:15 So we've got to take the derivative, denominator, using the quotient rule, times derivative of the numerator, which is just 2x, minus numerator, over derivative of the denominator, which is going to be 2x plus 1, over 2x plus 1.

03:55 Denominator squared.

04:05 Okay, simplifying the numerator, i get 2x squared plus 2x squared, plus 2x.

04:17 2x cubed plus 2x squared minus 4x minus, then in parentheses, 2x cubed plus x squared minus 2x and minus 1.

04:54 And the denominator is still the same thing.

05:07 2x cubed minus 2x cubed is 0...

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