Answer
(a) $f(0.9) \approx 4.8$
$f(1.1) \approx 5.2$
(b) A tangent line to a function that is concave down is above the function, so the approximations found in part (a) are too large.
Work Step by Step
(a) When $x = 1$:
$f(1) = 5$
$f'(1) = 2$
We can find the linear approximation at $a=1$:
$f(x) \approx f(a)+f'(a)(x-a)$
$f(x) \approx f(1)+f'(1)(x-1)$
$f(x) \approx 5+2(x-1)$
$f(x) \approx 3+2x$
We can approximate $f(0.9)$:
$f(x) \approx 3+2x$
$f(0.9) \approx 3+2(0.9)$
$f(0.9) \approx 4.8$
We can approximate $f(1.1)$:
$f(x) \approx 3+2x$
$f(1.1) \approx 3+2(1.1)$
$f(1.1) \approx 5.2$
(b) As $x$ increases, the value of $f'(x)$ decreases.
Therefore, $f(x)$ is concave down.
A tangent line to a function that is concave down is above the function, so the approximations found in part (a) are too large.
