Using mean value theorem show that: #x0#?

Question

Using mean value theorem show that:
#x< sin^-1x#, for #x>0#

Christopher Agresta · Accepted Answer

See below.

According to the mean value theorem, there exists #0 < zeta < x < 1# such that #(d/(dx)sin^-1(x))(zeta) = (sin^-1(x)-sin^-1(0))/(x-0) = (sin^-1(x))/x# but #(d/(dx)sin^-1(x))(zeta)=1/sqrt(1-zeta^2) > 1# finally # (sin^-1(x))/x = 1/sqrt(1-zeta^2) > 1# so # sin^-1(x) > x# or #x < sin^-1(x)#

Emilia Aguilar · Answer

undefined To show that $x < \sin^{-1}(x)$ for $x > 0$ using the Mean Value Theorem, we first define a function $f(x) = \sin^{-1}(x)$. Then, consider the interval $[0, x]$, where $x > 0$. By the Mean Value Theorem, there exists a point $c$ in the interval $(0, x)$ such that: $$f'(c) = \frac{f(x) - f(0)}{x - 0}$$ Where $f'(c)$ is the derivative of $f(x)$ at point $c$, and $f(x) = \sin^{-1}(x)$. The derivative of $\sin^{-1}(x)$ is $\frac{1}{\sqrt{1-x^2}}$. So, we have: $$\frac{1}{\sqrt{1-c^2}} = \frac{\sin^{-1}(x) - \sin^{-1}(0)}{x}$$ Simplify the above expression: $$\frac{1}{\sqrt{1-c^2}} = \frac{\sin^{-1}(x)}{x}$$ Now, since $0 < c < x$ and $x > 0$, we have $0 < c < 1$. Therefore, $1 - c^2 > 0$, so $\sqrt{1 - c^2}$ is real. Thus, we have: $$1 < x\sqrt{1 - c^2}$$ $$\frac{1}{\sqrt{1 - c^2}} < x$$ From the previous equation, we know that $\frac{1}{\sqrt{1 - c^2}} = \frac{\sin^{-1}(x)}{x}$. So, we get: $$\frac{\sin^{-1}(x)}{x} < x$$ And this implies: $$\sin^{-1}(x) < x$$ Therefore, $x < \sin^{-1}(x)$ for $x > 0$, as required.

Using mean value theorem show that: #x&lt; sin^-1x#, for #x&gt;0#?

Using mean value theorem show that: #x< sin^-1x#, for #x>0#

Using mean value theorem show that: #x< sin^-1x#, for #x>0#?

Using mean value theorem show that:
#x< sin^-1x#, for #x>0#