On algebra the pupils pass a subject of derivatives in the textbook of the 11th class. And here in this big paragraph it is featured for examination what is a tangent to the schedule and how to find and work out its equation.
Instruction
1. Let the y=f (x) function and a certain point of M with coordinates and and f(a) are given. And let it is known that there is f' (a). Let's work out the tangent equation. This equation as the equation of any other straight line which is not parallel to ordinate axis has y=kx+m appearance therefore for its drawing up it is necessary to find unknown k and m. With slope everything is clear. If the M belongs graphics and if from it it is possible to carry out the tangent not perpendicular to abscissa axis, then the slope of k is equal to f' (a). For calculation of unknown m we use that the required straight line passes through M. Sledovatelno's point if to substitute point coordinates in the straight line equation, then we will receive right equality of f(a) =ka+m. from here we find that m=f(a) - ka. It was necessary only to substitute values of coefficients in the equation direct.y=kx+my=kx+ (f(a) - ka) to y=f (a) +f' (a) (x-a) It follows from this that the equation has an appearance of y=f (a) +f' (a) (x-a).
2. To find the tangent equation to the schedule use a certain algorithm. First, designate also by a letter and. Secondly, calculate f(a). In the third, find a derivative from x and calculate f' (a). And at last, substitute found and, f(a) and f' (a) in a formula y=f (a) +f' (a) (x-a).
3. In order that it is better to understand how to use an algorithm, consider the following task. Work out the tangent equation for the y=1/x function in a point x =1. For the solution of this task use an algorithm of drawing up the equation. But at the same time consider that in this example f (x) function =2-h-h3, and =0.1 is given. In a statement of the problem the value of a point is specified and; 2. Therefore, f(a)=2-0-0=2;3. f' (x) =0-1-3х =-1-3х; f' ()=-1;4. Substitute the found numbers in the tangent equation to the schedule: y=f(a)+f' (a) (x-a) =2+(-1) (x-0) the =2nd. Answer: the y=2nd - x.