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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.

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What exactly should be two? At which point the value you're looking for? – Kaster Apr 25 '13 at 14:10
@Kaster Thanks, just took a practice test and now going over my incorrect answers. – yiyi Apr 25 '13 at 14:11
There's no way you can find it exactly, since you don't know if there are any other non-zero derivatives of the order higher than 2. All you can say is $$ f(7) \approx f(6) + f'(6)(7-6)+\frac 12 f''(6)(7-6)^2 = 3-\frac 12-1 = \frac 32 $$ I'm not sure, how you can get 2 exactly, since there's no more terms. – Kaster Apr 25 '13 at 14:18
any chance you overlooked something and $f''(6) = -1$? – Kaster Apr 25 '13 at 14:21
What do you mean by "2.5 is also possible answer"? Is it multiple choice test or something? – Kaster Apr 25 '13 at 14:23
up vote 1 down vote accepted

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 $\frac{-1}{2}$.

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$|f^{\prime\prime\prime} (x) | \leq M$ I get but why is $|M|\geq 2$?? – yiyi Apr 25 '13 at 15:26
Lagrange's Formula calls for $M$ to be the bound of $|f^{(n+1)}|$ on the interval $(a_0,x)$. That is actually the reason I considered only the first order Taylor Series, because then we can work out some information about the bound of $|f^{(n+1)}|$ because now $M$ is supposed to be a number such that $\forall x\in (6,7), \ |f^{\prime\prime}(x)| \leq M$ – David Blessing Apr 25 '13 at 15:30
$|M| \geq 2$ since $|f^{\prime\prime}(6)| = 2$ so, assuming this is a continuous second derivative, we know that on the interval $(6,7)$ the absolute value of the second derivative can get arbitrarily close to $2$, since $M$ is a bound of $|f^{\prime\prime}(x)|$ on $(6,7)$ we have that $M$ must be at least 2. #### edit: The assumption that $f^{\prime\prime}$ is continuous is justified since it is a requirement for any of this extrapolation to work (ie, the question doesn't make sense without it). – David Blessing Apr 25 '13 at 15:34

Since this is a multiple-choice question which only asks what value $f(7)$ could have, you would work to eliminate possibilities. The function has the value $f(6) = 3$ and the first derivative is $f(6) = -\frac{1}{2}$, so if there were no change in the slope over the interval ($f''(x) = 0$), the function would have $f(7) = 2.5$. However, the slope is decreasing at $x = 6$ ($f''(6) < 0$), so the slope will likely be more negative than $-\frac{1}{2}$ over the interval, leaving only the possibility of $f(7) = 2$ (A). The slope will not reverse itself and become positive anywhere on the interval to reach the values of 2.9, 3, or 4, because we are told that there are no inflection points in the graph of $f(x)$.

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In the exercise it is given that there are no points of inflection, i.e. $f''$ is always smaller than $0$. This means that $f(6+a) \leq f(6) + f'(6)*(a-6)$. In our cases this gives us $f(7)=f(6+1) \leq 3+ (-0.5)*1 = 2.5$ . This means that $2.5$ and $2$ are the only possible answers. But as $f''$ is smaller than $0$, the derivative is getting even smaller, therefore $2$ is the correct answer.

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Welcome to math.SE: I edited your answer, since it looked like a comment. Moreover, I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here. – A.P. Apr 25 '13 at 15:38

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