# 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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