The value of $\sqrt{24 . 99}$ is

$\sqrt{24.99}$ $\Delta y=f \left(x+\Delta x\right)-f\left(x\right)$ Take $ f\left(x\right)=\sqrt{x}$ $Let \,x = 25$ $\Delta x=-0.01$ $\therefore\, f\left(24.99\right)=\sqrt{25}+\Delta y$ $\approx5-0.001$ $\approx4.999$ $\Delta y \approx\frac{d y}{d x} \times\Delta x$ $\approx\frac{1}{2\sqrt{x}}\times-0.01$ $\approx-\frac{0.01}{10}$

The value of v24.99? is

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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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