Find the approximate value of f (2.01), where f (x) = 4x2+5x + 2
Find the approximate value of f (2.01), where f (x) = 4x2+5x + 2
Solution and Explanation
Let x = 2 and ∆x = 0.01. Then, we have: f(2.01) = f(x + ∆x) = 4(x + ∆x)2 + 5(x + ∆x) + 2 Now, ∆y = f(x + ∆x) − f(x)
∴ f(x + ∆x) = f(x) + ∆y
f(2.01)≈(4x2+5x+2)+(8x+5)∆x
28+0.21
=28.21
Hence, the approximate value of f (2.01) is 28.21.
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Notes on Applications 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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