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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$\begingroup$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 $f(x)$.
$\endgroup$ $\begingroup$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 $\frac{-1}{2}$.
$\endgroup$ 3 $\begingroup$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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