Question

Let $ f(x)={{x}^{3}}. $ Use mean value theorem to write $ \frac{f(x+h)-f(x)}{h}=f'(x+\theta h) $ , with $ 0<\theta <1 $ . If $ x\ne 0, $ then $ \underset{h\to 0}{\mathop{\lim }}\,\,\,\theta $ is equal to

• A. $ -1 $
• B. $ -0.5 $
• C. $ 0.5 $
• D. $ 1 $

Answer 1

Answer: (C)

Given, $ f(x)={{x}^{3}} $
$ \therefore $ $ f(x+h)={{(x+h)}^{3}} $
Now, $ f'(x)=3{{x}^{2}} $
$ \therefore $ $ f'(x+\theta h)=3{{(x+\theta h)}^{2}} $
Given, $ \frac{f(x+h)-f(x)}{h}=f'(x+\theta h) $
$ \Rightarrow $ $ \frac{{{(x+h)}^{3}}-{{x}^{3}}}{h}=3{{(x+\theta h)}^{2}} $
$ \Rightarrow $ $ \frac{{{x}^{3}}+{{h}^{3}}+3xh(x+h)-{{x}^{3}}}{h} $
$ =3({{x}^{2}}+{{\theta }^{2}}{{h}^{2}}+2x\theta h) $
$ \Rightarrow $ $ {{h}^{2}}+3{{x}^{2}}+3xh=3{{x}^{2}}+3{{\theta }^{2}}{{h}^{2}}+6x\theta h $
$ \Rightarrow $ $ 3x=3{{\theta }^{2}}h+6x\,\theta $
Taking limit on both sides, we get
$ \underset{h\to 0}{\mathop{\lim }}\,\,(h+3x)=\underset{h\to 0}{\mathop{\lim }}\,(3{{\theta }^{2}}h+6x\theta ) $
$ \Rightarrow $ $ 3x=6\,\,\underset{h\to 0}{\mathop{\lim }}\,\theta $
$ \Rightarrow $ $ \underset{h\to 0}{\mathop{\lim }}\,\theta =\frac{1}{2}=0.5 $