Saved Bookmarks
| 1. |
Draw the graph of f(x) = xcosx-sinx, x in [-3pi, 3pi] |
|
Answer» SOLUTION :`F(x) = xcosx-sinx` Clearly, domain is R. ALSO `f(x)` is non-periodic. `f^(')(x)=-xsinx` `f^(')(x) =0 rArr x=-3pi, -2pi, -pi, 0, pi, 2pi, 3pi` `f^(')(0^(-)) = (-)(-)(-) lt 0` and `f^(')(0^(+)) = (-)(+)(+) lt 0` So `x=0` is the POINT of inflection (as the derivative does not CHANGE sign in the neighbourhood of `x=0` Sign scheme of `f^(')(x)` is as follows:  Clearly, f is decreasing at `x=0` and has the point of minima at `x=pi, -2pi` and point of maxima at `x=-pi, 2pi` Since `f(x)` is an odd function, we check the graph for `x int [0, 3pi]` `f(0)=0, f(pi)=-pi` Thus, `f(x)` decreases from `0` to `-pi` in the interval `(0, pi)` Thus, `f(x)` increases from `-pi` to `2pi` in the interval `(pi, 2pi)` `f(3pi) =-3pi` Thus, `f(x)` decreases from `2pi` to `-3pi` in the interval `(2pi, 3pi)` From the above discussion the graph of `y=f(x)` is as shown in the following figure.
|
|