Simple Calculus AB MCQ Is my answer correct?
Let h be a differentiable function with a tangent line at x = π. The equation of the tangent line is y = −0.1x + 0.6. What must be true about h and h′ at x = π?
I. h(0) = 0.6
II. h(π) = 0.2858
III. h′(π) = 0.5
I and II
II and III
II only(My answer)
III only
 A: Yes, your answer is correct. We know that the value of the function at $\pi=0.2858$, and the slope of the tangent line at that value is $-0.1$, and so I and III are wrong. Therefore, II only is the correct answer.
A: If $f \colon \mathbb{R} \to \mathbb{R}$ is a differentiable function, we say that $\hat f_x(y) = f(x) + f'(x)(y-x)$ is the tangent line to $f$ at $x$, evaluated at $y$.
In your question, you have $\hat h_\pi(x) = -0.1 x + 0.6 = (0.6 - 0.1\pi) - 0.1(x - \pi)$, and from this we see immediately that $h(\pi) = 0.6 - 0.1 \pi$ and that $h'(\pi) = -0.1$.
From this we see:

*

*II is correct: indeed, $h(\pi) = 0.6 - 0.1\pi$.

*III is false: instead, $h'(\pi) = -0.1$.

It requires a little bit more work to show that I is false: consider
$$ 
h(x) = (x - \pi)^2 - \frac{0.1}{\pi} x+ 0.6 - 0.1\pi + 0.1$$
This function has $h(\pi) = 0.6-0.1\pi$, $h'(\pi) = -0.1$, and so it has the tangent line as given. However, by direct computation one checks that  $h(0) \neq 0.6$.
The intuition for I is that the tangent line only captures first-order information about $h$. So it only determines the slope and value at the point where the approximation is made (in this case $\pi$), and one cannot conclude anything more in general.
