According to the given plot, [tex]f'(1)=2[/tex]. Then the linear approximation to [tex]f(x)[/tex] at [tex]x=1[/tex] is
[tex]L(x)=f(1)+f'(1)(x-1)=6+2(x-1)=2x+4[/tex]
Then
[tex]f(0.95)\approx L(0.95)=5.9[/tex]
[tex]f(1.05)\approx L(1.05)=6.1[/tex]
On the interval [0, 1], the plot of [tex]f'(x)[/tex] is positive, so [tex]f(x)[/tex] is an increasing function here. But we can see that [tex]f'(x)[/tex] is approaching 0. This means tangent lines to [tex]f(x)[/tex] have a positive slope, but the slopes are approaching 0 and are thus becoming less steep. This in turn means the tangent lines lie above the curve, so the approximations are greater than the actual values of [tex]f(0.95)[/tex] and [tex]f(1.05)[/tex].
