선형근사와 미분소

(a) Since $e$ raised to any power is always positive (and thus never $0$), it follows that the only value that makes $f'(x) = (x-1)e^{-x^2} = 0$ is $x=1$. This is confirmed by plotting $f'$ where we only see the graph cross the $x$-axis at $x=1$ (though we do see the graph approach the $x$-axis) as $x$ increases or decreases without bound). (b) We know that $f''(x) > 0$ wherever $f'$ is increasing. From the graph of $f'$, we see that $f'$ is increasing approximately on the interval $-0.37 < x < 1.37$, and thus $f$ is concave up on this interval. (c) The tangent line approximation at $a = 2$ is $L(x) = f(2) + f'(2)(x-2) \approx -3 + 0.01832(x-2)$. Thus, $f(1.88) \approx L(1.88) \approx -3 + 0.01832(1.88-2) = -3.00022$. Since $f''(x) < 0$ at and around $x = 2$, $f$ is concave down there, and therefore the tangent line lies above the curve. This makes our estimate larger than the true value of $f(1.88)$.