Direct y=f(x) will be a tangent to the schedule represented in the drawing in a point h0 in case it passes through a point with coordinates (h0; f(x0)) also has slope of f' (x0). To find such coefficient, knowing features of a tangent, simply.
It is required to you
- - mathematical reference book;
- - simple pencil;
- - notebook;
- - protractor;
- - compasses;
- - handle.
Instruction
1. Pay attention that the schedule of the f (x) function differentiated in a point h0 differs in nothing from a tangent piece. So, it is rather close to a piece of l which passes through points (h0; f(х0)) and (х0+Δx; f (x0 + Δx)). To set a straight line which passes through a certain point And with coefficients (h0; f(х0)), it is necessary to specify its slope. At the same time the slope is equal Δy/Δx to a secant of a tangent (Δх→0) and aspires to number f‘ (x0).
2. If the value f‘ (x0) does not exist, then or there is no tangent, or it passes vertically. So, existence of derivative function in a point h0 is caused by existence of the not vertical tangent adjoining to a function graph in a point (h0, f(х0)). In this case the slope of a tangent will be equal to f' (h0). Thus, the geometrical meaning of a derivative – calculation of slope of a tangent becomes clear.
3. Represent in the drawing additional tangents which would adjoin to a function graph in x1 points, h2 and h3 and also note the corners formed by these tangents with abscissa axis (such corner is counted in the positive direction from an axis to a tangent straight line). For example, the first corner, that is, α1, will be sharp, the second (α2) – stupid, and the third (α3) is equal to zero as the drawn tangent straight line is parallel to an axis OH. In that case the tangent of an obtuse angle – negative value, a tangent of an acute angle – positive, and at tg0 result is equal to zero.