Find the maximum value of $f(x) = sin(sinx)$ for all $x \in R$

We have $f(x) = sin\,(sin\,x)$ , $x \in R$ Now, $-1 \le sin \,x \le 1$ for all $x \in R$ $\Rightarrow sin \left(-1\right) \le sin\left(sinx\right) \le sin \,1$ for all $x \in R$ [ $\because sin\, x$ is an increasing function on $\left[-1,1 \right]$ ] $\Rightarrow - sin\, 1 \le f\left(x\right) \le sin\, 1$ for all $x \in R$ This shows that the maximum value of $f\left(x\right)$ is $sin \,1$ .

Find the maximum value of $f(x) = sin(sinx)$ for a

Notes on Application of derivatives

Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by

$\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}$

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

If for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≤ f(x₂); then the function f(x) is known as increasing in this interval.
Likewise, if for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≥ f(x₂); then the function f(x) is known as decreasing in this interval.
The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x₁) < f(x₂) for strictly increasing and f(x₁) > f(x₂) for strictly decreasing.

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