how to find the value of a function from the first and second derivative. 
The function $f$ is twice differentiable, and the graph of $f$ has no points of inflection. if $f\left(6\right)=3,\, f^{\prime}\left(6\right) = -1/2,$ and $f^{\prime\prime}\left(6\right) = -2$  Which of the following could be the value of  $f\left(7\right)$?
(A) 2     (B) 2.5    (C) 2.9   (D) 3    (E) 4

From the answer sheet, I know that the answer should be two; however, I am unable to figure out why.
I have tried $y=f\left(a\right)+f^{\prime}\left(a\right)\left(x-a\right)$ but that gives the wrong answer.  I also tried to approximate with making a taylor series; but that failed horribly.
I know that the value must be less than $f\left(6\right)$ because the tangent is negative and the second derivative is negative as well.

copied the question exactly from the pdf.

If we use a truncated taylor series, the answer does indeed work out:
$$
f(7) \approx f(6) + f^\prime (6)(7-6) + \frac{1}{2} f^{\prime\prime} (6) (7-6)^2 = 3 - \frac{1}{2} + \frac{1}{2} \cdot (-2) = \frac{3}{2}
$$
We now need a bound on our error, using Lagrange Error Bound:
$$
|E_n (x)| \leq \frac{M}{(n+1)!} |x-a_0|^{n+1} = \frac{M}{(3)!} |x-6|^{3}
$$
Where $M$ is some value satisfying $|f^{(n+1)}| = |f^{\prime\prime\prime} (x) | \leq M$ on the interval $(6,x)$.
Since we don't know the value of $f^{\prime\prime\prime}$, we can sloppily just assume the closest answer is correct.
Alternatively, we truncate earlier:
$$
f(7) \approx f(6) + f^\prime (6)(7-6) \pm \frac{M}{2!} |7-6|^{2} = 3 - \frac{1}{2} + \frac{M}{2} = \frac{5}{2} \pm \frac{M}{2}
$$
Recalling that $f^{\prime\prime}(6) = -2$, we know that $|M| \geq 2$, which creates a lower bound on $|M|$.
So, any value $y \in \mathbb{R}$ that satisfies this equation is a possible answer,
$$
\frac{3}{2} = \frac{5}{2} - 1 \leq y \leq \frac{5}{2} + 1 = \frac{7}{2}
$$
So, the best possible answer is indeed $2$, however, $2.5$ is also within the tolerance.
edit: Added more about definition of Lagrange's Error Bound
edit: I now see that I did not read fully the question, since there are no inflection points, we can modify our upper bound to reflect that the derivative is always decreasing, so $\frac{3}{2} \leq y \leq \frac{5}{2}$ since there is no way for the derivative to be greater than $\frac{-1}{2}$.
