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.
Top Questions on Applications of Derivatives
- The total revenue in Rupees received from the sale of \(x\) units of a product is given by \(R(x)=13x^2+26x+15\).The marginal revenue, when \(x=7\) is
- CBSE CLASS XII
- Mathematics
- Applications of Derivatives
If f (x) = 3x2+15x+5, then the approximate value of f (3.02) is
- CBSE CLASS XII
- Mathematics
- Applications of Derivatives
- Show that the function given by f(x)=\(\frac{logx}{x}\) has maximum at x=e
- Mathematics
- Applications of Derivatives
It is given that at x = 1, the function x4−62x2+ax+9 attains its maximum value, on the interval [0, 2]. Find the value of a.
- CBSE CLASS XII
- Mathematics
- Applications of Derivatives
Find the maximum profit that a company can make, if the profit function is given by p(x) = 41−24x−18x2
- CBSE CLASS XII
- Mathematics
- Applications of Derivatives
Questions Asked in CBSE CLASS XII exam
- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
- For what values of x,\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)=O?
- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
- Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
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.
Read More: Application of Derivatives