Solution (a) Since f′(x)=cosx, then f(0)=sin0=0 and f′(0)=cos0=1. Using Equation (3), the linear approximation L(x) to f at 0 is L(x)=f(0)+f′(0)(x−0)=x
So, for x close to 0, the function f(x)=sinx can be approximated by the line L(x)=x.
(b) The approximate values of sinx using L(x)=x, the true values of sinx, and the absolute error in using the approximation are given in Table 3. From Table 3, we see that the further x is from 0, the worse the line L(x)=x approximates f(x)=sinx.
L(x)=x | f(x)=sinx | Error: |x−sinx| |
0.1 | 0.0998 | 0.0002 |
−0.3 | −0.2955 | 0.0045 |
0.4 | 0.3894 | 0.0106 |
0.5 | 0.4794 | 0.0206 |
π4≈0.7854 | 0.7071 | 0.0783 |
(c) See Figure 11 for the graphs of f(x)=sinx and L(x)=x.