Find the slope of a line tangent to the curve of the given equation at the given point. Sketch t

Find the slope of a line tangent to the curve of the given...

Key Concepts

Graphing Functions

Graphing functions involves plotting the set of all points that satisfy the function's equation on a coordinate plane. This visual representation helps in understanding the overall behavior of the function, identifying key features such as intercepts, maxima, minima, and the manner in which the tangent lines interact with the curve.

Slope

The slope is a measure of the steepness and direction of a line, defined as the ratio of the vertical change to the horizontal change between two points. In the context of calculus, the slope of the tangent line at a point on a curve is given by the derivative of the function at that point.

Derivative

The derivative of a function is a fundamental concept in calculus that represents the rate at which the function's value changes with respect to changes in its input. It is used to compute the slope of the tangent line to the function at any given point, providing insight into the behavior and instantaneous rate of change of the function.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve precisely at that point and has the same slope as the curve at that point. It serves as a local linear approximation of the curve and is essential for understanding how the function behaves near that specific point.

Transcript

0:00 Welcome to numerid.

00:01 In the current problem, we are about to find the derivative of the given curve y equals to 2x minus x squared.

00:19 And we also have to find the slope or the tangent.

00:25 Okay, we have to find the slope of the tangent at the given point.

00:30 Minus 1 minus 3 and also we have to sketch this.

00:36 So with no further delay we will start by defining this as fx.

00:44 Now for definition we know limit h tends to 0 f of x plus h minus f of x divided by h.

00:59 Correct so we have to start by finding f of x plus h which is 2 into x plus h minus x plus h whole square which is equals to 2x plus 2h minus x square minus 2xxxxxx minus h minus h therefore, f of x plus h minus f of x would be 2x plus 2h minus x squared minus 2xx minus 2x minus h squared minus 2x minus h squared.

01:48 Minus this entire 3 .2x plus x minus x squared.

01:56 So this will be minus 2x plus x squared.

01:59 So this term, this term and this term and this term would cancel.

02:07 So we would be left with 2h minus 2 x h minus h square.

02:16 Taking h common we would have 2 minus 2x minus h.

02:23 Therefore then if we divide the left -hand equation by h that is fx plus h minus fx by h we will also be having 2 minus 2x minus h as if that h and that h comes in the denominator that h and divided by h cancels out each other but now if we find what is d by d x so d by d x is equal to limit h tends to 0, f of x plus h minus, that is this expression, f of x divided by h.

03:12 That is, limit h tends to 0, we write the entire expression.

03:22 Now, we will write it by term by term limits.

03:29 Limit of a constant minus limit of a, what should i say? yes, because when we are taking limits with respect to the variable h, everything else is a constant.

03:43 So, wherever we get h, we will substitute those places by h becoming negligibly small, which is about to be zero.

03:51 So, here h have no role to play, so we get 2 minus.

03:56 Here also h has no role to play hence to x, but this term will vanish because as h tends to 0, this entire expression tends to 0.

04:06 So right doesn't change in everything.

04:10 Now, with this understanding is the derivative at point.

04:18 So this is the expression for d .y dx and we have to also find d y dx at the point next 1 minus 3.

04:36 Okay, so let us first write a table.

04:43 So here are on x values here are with y values and i generally start by minus 3 minus minus 2 minus 1, 0, 1, 2, 3.

04:56 And we have to evaluate the expression, that is y is equal to 2x minus x square.

05:06 So if i put 0, let's substitute 0...

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