Find the approximate value of f(3.02), where f(x) = \(3x^2 + 5x + 3\).

Solution :

Let y = f(x), x = 3 and x + \(\delta x\). Then, \(\delta x\).= 0.02.

For x = 3, we get

y = f(3) = 45

Now, y = f(x) \(\implies\) y = \(3x^2 + 5x + 3\)

\(\implies\) \(dy\over dx\) = 6x + 5 \(\implies\)  \(({dy\over dx})_{x = 3}\) = 23

Let \(\delta y\) be the change in y due to change \(\delta x\) in x. Then,

\(\delta y\) = \(dy\over dx\) \(\delta x\)  \(\implies\)  \(\delta y\) = \(23 \times 0.02\) = 0.46

f(3.02) = y + \(\delta y\) = 45 + 0.46 = 45.46


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