$\mathbf{Question:}$ Find the values $a$ and $b$ such that the function is differentiable at $x=0$

$$ f(x)= \begin{cases} x^{2}+1 &x≥0\\ a\sin x+b\cos x & x<0\\ \end{cases} $$


$f(x)$ is differentiable at $x=0$ if $f'(0)$ exists. This implies that for $f$ to be differentiable at $x=0$, the left hand limit and the right hand limit must exist and be equal.

$$ \begin{align} \lim_{x\to 0-}f'(0) & =\lim_{x\to 0-}\frac{f(x)-f(0)}{x} \\ & =\lim_{x\to 0-}\frac{a\sin x +b\cos x-1}{x} \\ & =\lim_{x\to 0-}\frac{a\sin x}{x}+\frac{b\cos x-1}{x}=a \\ \end{align} $$

$$ \begin{align} \lim_{x\to 0+}f'(0) & =\lim_{x\to 0+}\frac{f(x)-f(0)}{x} \\ & =\lim_{x\to 0+}\frac{x^{2}+1-1}{x} = 0 \end{align} $$

Therefore, $a=0$

To find $b$, we can use the fact that if $f(x)$ is differentiable at $x=0$ then, it must be continuous at $x=0$.

So if $f(x)$ is continuous, $\lim_{x \to0-}f(x) = \lim_{x \to0+}f(x)=b$

$$ \begin{align} \lim_{x\to 0-}f(x) & =\lim_{x\to 0-}a\sin x +b\cos x \\ & = a\sin (0) + b\cos (0) = b \end{align} $$

$$ \begin{align} \lim_{x \to 0+}f(x) & = \lim_{x \to0+}x^{2}+1 =1 \end{align} $$

Therefore, $b=1$

Thus, $ f(x)= \begin{cases} x^{2}+1 &x≥0\\ \cos x & x<0\\ \end{cases} $ is differentiable at $x=0$

  • 1
    $\begingroup$ How do you know that $\lim_{x\to 0-}\frac{b\cos x-1}{x}=0$ without knowing that $b=1$? Outside of this, your analysis looks good. $\endgroup$ – Axion004 Jun 9 '20 at 4:26

$$f(0)=1$$ $$\lim_{x\to 0^+}f(x)=\lim_{x\to0^+}(a\sin(x)+b\cos(x))=b$$ $f$ is continuous at $x=0$ if $b=f(0)=1$.

$$f'(0^-)=\lim_{x\to 0^-}\frac{f(x)-f(0)}{x-0}$$ $$=\lim_{x\to0^-}x=\color{red}{0}$$ $f$ is differentiable at $x=0$ if $$\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\color{red}{0}$$ or $$\lim_{x\to0^+}\frac{a\sin(x)+\cos(x)-1}{x}=0$$ but $$\sin(x)\sim x \; and \; \cos(x)-1\sim \frac{-x^2}{2}$$ thus $a=0$.


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