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.