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 $f(1.05)$.