For \(a \in \mathbb{R}\), consider the real valued function defined on \((-1,1)\):
\[
f(x)=
\begin{cases}
\dfrac{(1+x)^{1/3}-(1+2x)^{1/4}}{x}, & x\neq0,\\
a,& x=0.
\end{cases}
\]
If \(f\) is differentiable at \(x=0\), then value of \(a+f'(0)\) is:
Show Hint
For piecewise functions involving powers, first enforce continuity to determine the unknown constant. Then use Taylor/Binomial expansion to obtain the derivative quickly.
Step 1: Spot the indeterminate form. As $x \to 0$ the numerator $(1+x)^{1/3}-(1+2x)^{1/4}$ becomes $1-1=0$ and the denominator $x$ becomes $0$, so we have a clean $\frac{0}{0}$ form. That is exactly the situation where a short Taylor expansion of each root works beautifully. Step 2: Expand the first root to two useful terms. Using $(1+t)^n = 1 + nt + \frac{n(n-1)}{2}t^2 + \cdots$ with $n=\frac13$, $t=x$: \[ (1+x)^{1/3} = 1 + \frac{x}{3} - \frac{x^2}{9} + \cdots \] Step 3: Expand the second root the same way. With $n=\frac14$, $t=2x$: \[ (1+2x)^{1/4} = 1 + \frac{2x}{4} + \frac{\frac14\left(-\frac34\right)}{2}(2x)^2 + \cdots = 1 + \frac{x}{2} - \frac{3x^2}{8} + \cdots \] Step 4: Subtract and divide by $x$. The constant terms cancel: \[ (1+x)^{1/3}-(1+2x)^{1/4} = \left(\frac13-\frac12\right)x + \left(-\frac19+\frac38\right)x^2 + \cdots = -\frac{x}{6} + \frac{19}{72}x^2 + \cdots \] Dividing by $x$ gives $f(x) = -\frac16 + \frac{19}{72}x + \cdots$ Step 5: Read off $a$ from continuity. For $f$ to be even continuous (a must for differentiability) we need $a = \lim_{x\to0} f(x) = -\frac16$. Step 6: Read off $f'(0)$ from the linear coefficient. Since $f(x) = -\frac16 + \frac{19}{72}x + \cdots$ is already in power-series form, the coefficient of $x$ is $f'(0) = \frac{19}{72}$. Step 7: Add them. \[ a + f'(0) = -\frac16 + \frac{19}{72} = -\frac{12}{72} + \frac{19}{72} = \frac{7}{72} \] \[ \boxed{\dfrac{7}{72}} \]