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Find the approximate value of f(3.02), where {f{{({x})}}}={3}{

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

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Benjamin Discussed

Find the approximate value of f(3.02), where

f (x) = 3 x^{2} + 5 x + 3

11 mins ago

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11 mins ago

Text solutionVerified

Solution: Here, we will use the definition of derivatives:

f^{'} (x) = \frac{f ( x + Δ x ) - f ( x )}{( x + Δ x ) - x}

f^{'} (x) = \frac{f ( x + Δ x ) - f ( x )}{Δ x}

Here,

x = 3 and Δ x = 0.02

∴ f^{'} (3) = \frac{f ( 3 + 0.02 ) - f ( 3 )}{0.02} \to (1)

Here,

f^{'} (x) = 6 x + 5

∴ f^{'} (3) = 6 (3) + 5 = 23

f (3) = 3 (9) + 5 (3) + 3 = 45

So,putting these values in (1),

23 = \frac{f ( 3.02 ) - 45}{0.02}

0.46 + 45 = f (3.02)

\Rightarrow f (3.02) = 45.46

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Question Text	Find the approximate value of f(3.02), where $f (x) = 3 x^{2} + 5 x + 3$ .
Answer Type	Text solution:1
Upvotes	150