If f(x)= {-5x+5x^2 -1<x<0
{sin(cx) 0<=x<=2ð
To check if f(x) is differentiable at 0, we need to evaluate lim Q(h) when h->0^-.
Find lim Q(h) when h->0^-.
To find the limit of Q(h) as h approaches 0 from the left side, we need to plug in the expression for f(x) and evaluate it as h approaches 0 from the negative side.
First, let's recall the definition of the limit:
lim Q(h) as h->0^- = lim f(x) as x->0^-,
Since h = x - 0, we can rewrite the above expression as:
lim Q(h) as h->0^- = lim f(x) as x->0^-.
Now, let's evaluate the limit for f(x) as x approaches 0 from the negative side.
Since -1 < x < 0 in the first part of the function, we can substitute x = 0 in that expression:
f(x) = -5x + 5x^2.
f(0) = -5(0) + 5(0)^2
= 0.
So, when x approaches 0 from the negative side, f(x) approaches 0.
Therefore, the limit of Q(h) as h approaches 0 from the negative side is:
lim Q(h) as h->0^- = lim f(x) as x->0^- = 0.