Question

If $f\left(x\right)=x^3$ use mean value theorem $\frac{f\left(x+h\right)-f\left(x\right)}{h}=f\text{'}\left(x+\theta h\right)$ with $0<\theta <1$ and if $x\ne 0$, then $\underset{h\to 0}{\lim }\theta$ is equal to

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

Answer 1

The correct option is C

$0.5$

Explanation for the correct option:

Step 1. Form the equation:

For the function $f\left(x\right)=x^3$, the derivative is given as: $f\text{'}\left(x\right)=3x^2$.

Now, the mean value theorem $\frac{f\left(x+h\right)-f\left(x\right)}{h}=f\text{'}\left(x+\theta h\right)$ can be applied as:

$$\begin{gathered} \frac{{\left(x+h\right)}^3-x^3}{h}=3{\left(x+\theta h\right)}^2 \\ \Rightarrow \frac{x^3+3x^{2}h+3x{h}^2+h^3-x^3}{h}=3\left(x^2+2x\theta h+{\theta }^2{h}^2\right) \\ \Rightarrow 3x^2+3xh+h^2=3x^2+6x\theta h+3{\theta }^2{h}^2 \\ \Rightarrow h\left(3x+h\right)=h\left(6x\theta +3{\theta }^{2}h\right) \\ \Rightarrow 3x+h=6x\theta +3{\theta }^{2}h \end{gathered}$$

Step 2. Find the value of $\underset{h\to 0}{\lim }\theta$.

In the equation $3x+h=6x\theta +3{\theta }^{2}h$, take $\underset{h\to 0}{\lim }$ on both sides and find the value of $\underset{h\to 0}{\lim }\theta$.

$$\begin{gathered} \underset{h\to 0}{\lim }\left(3x+h\right)=\underset{h\to 0}{\lim }\left(6x\theta +3{\theta }^{2}h\right) \\ \Rightarrow 3x=6x\underset{h\to 0}{\lim }\theta +0 \\ \Rightarrow \frac{3x}{6x}=\underset{h\to 0}{\lim }\theta \\ \Rightarrow 0.5=\underset{h\to 0}{\lim }\theta \end{gathered}$$

Thus the value of $\underset{h\to 0}{\lim }\theta$ is $0.5$.

Hence, the correct option is C.