Let f(x)=x3. Use the mean value theorem to write f(x+h)-f(x)h=f'(x+θh) with 0<θ<1. If x≠0,then limhh→0θ= ?
-1
-0.5
0.5
1
Explanation for the correct option:
Given that: f(x)=x3
Therefore, f(x+h)=f(x)+hf'(x)+h22f''(x+θ1h), where 0<θ<1
Applying Lagrange's theorem,
f(x+h)-f(x)=hf'(x+θh),0<θ<1f'(x+θh)-f'(x)=θ1hf''(x+θ2h)θhf''(x+θ2θh)=h2f'(x+θ1h)θ=12f''(x+θ1h)f'(x+θ2h)
now,
=limh→0θ=12f''(x)f''(x)=12
f'' is continuous in x,x+h
Hence, the correct option is (C).
If fx=x3 use mean value theorem fx+h-fxh=f'x+θh with 0<θ<1 and if x≠0, then limh→0θ is equal to