If \(f(x)=\begin{cases}ax+b & : x\le -1\\ ax^3+x+2b & : x>-1\end{cases}\) is differentiable for \(x\in \mathbb{R}\), find (a, b)
(-1/2, 1)
Differentiability at \(x=-1\) first requires continuity there: \(a(-1)+b=a(-1)^3+(-1)+2b\), i.e. \(-a+b=-a-1+2b\).
This simplifies to \(b=-1+2b\), so \(b=1\).
Next, the left-hand derivative of \(ax+b\) is the constant \(a\), while the right-hand derivative of \(ax^3+x+2b\) is \(3ax^2+1\), which equals \(3a+1\) at \(x=-1\).
Equating the two derivatives: \(a=3a+1\Rightarrow -2a=1\Rightarrow a=-\dfrac12\).
Therefore \((a,b)=\left(-\dfrac12,\,1\right)\).
Similar Questions
If f(x) is differentiable at x = a and f'(a) = 0, then find the value of
\(\displaystyle\lim_{h\to 0}\frac{f(a+2h^2)-f(a-2h^2)}{h^2}\)
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