# Interval for Lagrange remainder in Taylor series

I have a question about Taylor series with Lagrange remainder. In my example I have point $$x_1 = x_0 + \frac{3h}{4}$$ where h is constant. Now, it is asked to write function $$y(x_0+h)$$ in Taylor series with Lagrange remainder using point $$x_1$$ where remainder is of order 2(second derivative). Here is the solution: $$y(x_0+h) = y(x_1+\frac{h}{4}) = y(x_1)+y'(x_1)\frac{h}{4}+\frac{y''(\alpha)}{2!}\frac{h^2}{16}$$ This is get using that $$x_0 = x_1-\frac{3h}{4}$$. Now, in solution it is also written that $$\alpha\in(x_0,x_0+h)$$ and that is part I don't understand. Using formula for Lagrange remainder $$\alpha$$ should belong to $$(x_1,x_1+\frac{h}{4})$$ or in terms of $$x_0$$ $$(x_0+\frac{3h}{4},x_0+h)$$. So, my right bound of interval is same as in the solution, but I don't understand why they set left bound to $$x_0$$. Any help is appreciated.

You are correct. Anyway, if $$h>0$$ then $$\alpha\in(x_1,x_1+\frac{h}{4})=(x_0+\frac{3h}{4},x_0+h)\subset (x_0,x_0+h)$$ so, it is also true that $$\alpha$$ belongs to the interval $$(x_0,x_0+h)$$.