Question

Let $f(x)=x^{3}$. Use the mean value theorem to write $[f(x+h)-f$ $(x)] / h=f^{\prime}(x+\theta h)$ with $0<\theta<1$. If $x \neq 0$, then $\lim _{h \rightarrow 0} \theta=$
A. $-1$
B. $-0.5$
C. $0.5$
D. 1

Answer 1

Hint: In this question, we used the mean value theorem.So first we differentiate the given equation, then we use the mean value theorem and solving it further by taking the limit as h approaches to 0, we find the desirable answer.

Formula Used:
Mean value theorem:
if a function f is continuous on the closed interval $[a,b]$ and differentiable on the open interval (a,b), then there exists a point $c$ in the interval $[a,b]$ such that $f'(c)$ equals the function's average rate of change throughout $[a,b]$.

Complete Step by step Solution:
Given, $f(x)=x^{3}$
$\therefore f(x+h)=(x+h)^{3}$
The instantaneous rate of change of a function at a certain position is known as the derivative of a function. The derivative provides the precise slope of the curve at a given location. The function's derivative, or the derivative of $y$ with respect to the variable $x$, is denoted by the symbol $dy/dx$.
The derivative of the function is
Now, $f^{\prime}(x)=3 x^{2}$
$\therefore f^{\prime}(x+\theta h)=3(x+\theta h)^{2}$
Given that
$\text{ }\dfrac{f(x+h)-f(x)}{h}={{f}^{\prime }}(x+\theta h)$
$\Rightarrow \dfrac{(x+h)^{3}-x^{3}}{h}=3(x+\theta h)^{2}$
Simplifying the expression we get
$\Rightarrow \dfrac{x^{3}+h^{3}+3 x h(x+h)-x^{3}}{h}$$=3\left(x^{2}+\theta^{2} h^{2}+2 x \theta h\right)$
$\Rightarrow h^{2}+3 x^{2}+3 x h=3 x^{2}+3 \theta^{2} h^{2}+6 x \theta h$
${{h}^{2}}+3xh=3{{\theta }^{2}}{{h}^{2}}+6x\theta h$
 Taking limit on both sides, we get
$\lim _{h \rightarrow 0}(h+3 x)=\lim _{h \rightarrow 0}\left(3 \theta^{2} h+6 x \theta\right)$
The limit of a function is denoted by the shorthand notation $\lim _{x \rightarrow p} f(x)=L$, which shows that as $x$ approaches a from both the left and the right of $x=a$, the output value approaches $L$.
$3x=0 + 6x. \underset{h\to 0}{\mathop{\lim }}\,\theta $
$\Rightarrow \lim _{h \rightarrow 0} \theta=\dfrac{1}{2}=0.5$
So the correct answer is option (C).

Note: The Mean Value Theorem is a specific case of Rolle's Theorem. We take into account differentiable functions $f$ in Rolle's theorem that have zero ends. Rolle's theorem is generalized by the mean value theorem by taking into account functions whose ends are not always zero.
The mean value theorem is a key finding in real analysis that is highly helpful for understanding how functions behave in higher mathematics. Before developing it logically as a general case of Rolle's Theorem and comprehending its significance, we will simply state the theorem.