Question

I really dont know how to start this Calculus 1 question!

Answer 1

According to the given plot,
`f'(1)=2`. Then the linear approximation to
`f(x)` at
`x=1` is

`L(x)=f(1)+f'(1)(x-1)=6+2(x-1)=2x+4`

Then

`f(0.95)\approx L(0.95)=5.9`

`f(1.05)\approx L(1.05)=6.1`

On the interval [0, 1], the plot of
`f'(x)` is positive, so
`f(x)` is an increasing function here. But we can see that
`f'(x)` is approaching 0. This means tangent lines to
`f(x)` 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
`f(0.95)` and
`f(1.05)`.