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The Problem:

Find the value of constants $c_1, c_2, c_3, c_4$ for which function $f: \mathbb{R} \rightarrow \mathbb{R} $ (written below) is differentiable:

$$ f(x) = \begin{cases} c_1 \arctan x + c_2 \cos x + \sin x & \quad x < 0\\ \ln(1+x^2) - x \ln 2 + 2^x & \quad 0 \leq x \leq 1\\ c_3 (x-1)(x-2)\cdots (x-2016) + c_4 (x+1) & \quad 1 < x \\ \end{cases} $$

Potential solution:

First I need to find values for which the function is continuous.

$\lim_{x \rightarrow 0^-}$ of 1st case should be equal to the 2nd case evaluated at $x=0$.

$$\lim_{x \rightarrow 0^-} c_1 \arctan x + c_2 \cos x + \sin x = \ln 1 + 1$$ $$\Rightarrow c_2 = 1 ,\ c_1 \in \mathbb{R}$$

2nd case evaluated at $x=1$ should be equal to $\lim_{x \rightarrow 1^+}$ of the 3rd case.

$$\ln 2 - \ln2 + 2 = \lim_{x \rightarrow 1^+} c_3 (x-1)(x-2)\cdots (x-2016) + c_4 (x+1)$$

$$\Rightarrow c_4 = 1 ,\ c_3 \in \mathbb{R}$$

Solving finds the values of $c_2$ and $c_4$ for which the function is continuous.

To find values for which it is differentiable I first need to find derivatives of each of the parts and then go through the previous process again. But, I don't know how to find the derivative of the 3rd case. Here's what I tried.

The derivative:

$$ f'(x) = \begin{cases} c_1 \frac{1}{1+x^2} - c_2 \cdot \sin x + \cos x & \quad x < 0\\ 2x \cdot \frac{1}{1+x^2} - \ln 2 + \ln 2 \cdot 2^x & \quad 0 \leq x \leq 1\\ ???\ \ c_3 (2016 x^{2015} + \ldots) + c_4 \ \ ??? & \quad 1 < x \\ \end{cases} $$

Repeating the same process with the derivative instead of original function:

$$\lim_{x \rightarrow 0^-} c_1 \frac{1}{1+x^2} - 1 \cdot \sin x + \cos x = 2\cdot0\frac{1}{1+0} - \ln2 + 2^0\ln 2$$

$$\Rightarrow c_1 = -1$$

Here's the problematic part: $$ 2 \cdot \frac{1}{1+1^2} - \ln 2 + \ln 2 \cdot 2^1 = \lim_{x \rightarrow 1^+} c_3 (2016 x^{2015} + \ldots) + 1$$

$$c_3 = ???$$

Any ideas on how to find this derivative and then its limit? Is there a different approach for solving this problem?

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$\lim_\limits{x\to 1^+} f'(x)$

We don't need to find the exact derivative of $c_3(x−1)(x−2)⋯(x−2016)+c_4(x+1)$ When we differentiate we get $ c_3 (x-2)\cdots(x-2016) + c_3(x-1) [(x-3)\cdots(x-2016) + (x-2)(x-4)\cdots]+c_4$ And as $x$ approaches $1, c_3(x-1)[\text{everything inside the brackets}] = 0$

leaving $c_3 (2015!) + c_4$

regarding your work on $c_1$

you calculated the derivatives correctly, but you made an error when you evaluated them in a neighborhood of 0.

$\lim_\limits{x=0^+} f'(x) = 2\cdot0\frac{1}{1+0} - \ln2 + 2^0\ln 2 = 0$

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  • $\begingroup$ Ahhh, I see. That's a really nice solution. Thank you very much! I edited my post to include your correction for $c_1$. $\endgroup$ – Radiant Aug 30 '16 at 23:33
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The derivative of $c_3(x-1)(x-2)\cdots (x-2016) + c_4(x+1)$ is given by

$$ c_3 \sum_{i=1}^{2016} \prod_{\substack{j=1 \\ j \neq i}}^{2016} (x-j) + c_4. $$

Hence the limit as $x \to 1+$ is given by

$$ c_3\prod_{j=2}^{2016} (1-j) + c_4. $$

Now you can equate the limits as you were doing to deduce the constants $c_3$ and $c_4$.

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