Chapter 2: Q 34 (page 238)

Find a function f that has the given derivative $f'$ and value f(c).
$f' (x) = \frac{1}{1 + 4 x^{2}}, f (0) = 1$

Short Answer

Expert verified

The anti derivative is $f (x) = \frac{1}{2} \tan^{- 1} (2 x) + 1$

Step by step solution

Step 1:Given information

The given expression is $f' (x) = \frac{1}{1 + 4 x^{2}}, f (0) = 1$

Step 2:Simplification

Given $f' (x) = \frac{1}{1 + 4 x^{2}}$

We will use an intelligent guess-and-check method to find f

width="198" height="44" role="math">⇒ddxtan-12x=21+2x2

. We need only divide our guess by that constant. Try the function $f (x) = \frac{1}{2} \tan^{- 1} 2 x$

$\Rightarrow \frac{d}{d x} (\frac{1}{2} \tan^{- 1} 2 x) = \frac{1}{2} . (\frac{2}{1 + 4 x^{2}})$

$= \frac{1}{1 + 4 x^{2}}$

Hence,

$f (x) = \frac{1}{2} \tan^{- 1} (2 x)$ $+ c$

Now,

$f (0) = 1 \Rightarrow 1 = \frac{1}{2} (\tan^{- 1} 0) + c \Rightarrow c = 1$

Therefore,

The antiderivative is $f (x) = \frac{1}{2} \tan^{- 1} (2 x) + 1$

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Find a function f that has the given derivative $f'$ and value f(c).
$f' (x) = \frac{1}{1 + 4 x^{2}}, f (0) = 1$

Short Answer

Step by step solution

Step 1:Given information

Step 2:Simplification

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Most popular questions from this chapter

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Find a function f that has the given derivative $f'$ and value f(c).
$f' (x) = \frac{1}{1 + 4 x^{2}}, f (0) = 1$