Question

Find the approximate value of f(2.01), where $f(x) = 4{x^2} + 5x + 2$.

Answer 1

Hint: You can find the approximate value of f(2.01) using approximation by differentials method given as $f(x + \Delta x) = y + \Delta y$ where $f(x) = y$ and $\Delta y = \dfrac{{dy}}{{dx}}\Delta x$. Substitute the values and find the value.

Complete step-by-step answer:

We are given a function and we need to find the approximate value of f(2.01). The function is given as follows:
$f(x) = 4{x^2} + 5x + 2$
Let us assume the function value is equal to the variable y. Then, we have:
$f(x) = y$
$y = 4{x^2} + 5x + 2...........(1)$
We can use the method of approximation by differentials to find the value of f(2.01).
The method of approximation for a function f(x) such that $f(x) = y$ with $\Delta y = \dfrac{{dy}}{{dx}}\Delta x$ is given as follows:
$f(x + \Delta x) = y + \Delta y..........(2)$
We take x as 2 and $\Delta x$ as 0.01. Hence, we have:
$x = 2$
$\Delta x = 0.01$
Now, let us calculate y at x = 2.
$y = f(2)$
We have as follows:
$y = 4{(2)^2} + 5(2) + 2$
Simplifying, we have:
$y = 16 + 10 + 2$
Adding the terms, we have:
$y = 28..........(3)$
We know the value of $\Delta y$ is given as follows:
$\Delta y = \dfrac{{dy}}{{dx}}\Delta x............(4)$
Differentiating equation (1), we have:
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(4{x^2} + 5x + 2)$
Simplifying, we have:
$\dfrac{{dy}}{{dx}} = 8x + 5$
Substituting the above expression in equation (4), we get:
$\Delta y = (8x + 5)\Delta x$
Substituting the value of x and $\Delta x$, we have as follows:
$\Delta y = (8(2) + 5)(0.01)$
Simplifying the above expression, we have:
$\Delta y = (16 + 5)(0.01)$
$\Delta y = (21)(0.01)$
$\Delta y = 0.21.............(5)$
Substituting equation (3) and equation (5) in equation (2), we obtain as follows:
$f(2 + 0.01) = 28 + 0.21$
Adding the terms, we get as follows:
$f(2.01) = 28.21$
Hence, the approximate value of f(2.01) is 28.21.

Note: Be careful when evaluating the term $\Delta y$, you might make a mistake by forgetting the $\Delta x$ factor and express it as $\Delta y = \dfrac{{dy}}{{dx}}$ which is wrong. The formula to calculate $\Delta y$ is $\Delta y = \dfrac{{dy}}{{dx}}\Delta x$. The exact value of the expression f(2.01) is 28.2104.