Question

4. Determine o valor da derivada da função f(x) = 42x + 3(2-x²)√4x+1 no ponto x=2

Answer 1

Resposta:

Seja a função $f$ assim definida:

$f(x) = 42x + 3(2 - x^2)\sqrt{4x + 1}.$

Determinemos, inicialmente, o domínio de $f:$

$4x + 1 \geq 0 \Longleftrightarrow 4x \geq -1\Longleftrightarrow x \geq -\frac{1}{4}.\\\\D_f = \left\{x \in \mathbb{R}\, |\, x \geq -\frac{1}{4} \right\}.$

Encontremos a derivada de $f:$

$f'(x) = 42 + 3 \cdot \left( 2 - x^2 \right) \cdot \frac{1}{2} \left(4x + 1 \right)^{ -\frac{1}{2}} \cdot 4 + 3 \cdot \left(-2x \right) \cdot \sqrt{4x + 1}\\\\\Longleftrightarrow f'(x) = 42 + 6 \cdot \frac{\left(2 - x^2 \right)}{\sqrt{4x + 1} } -6x \cdot \sqrt{4x + 1}\\\\\Longleftrightarrow f'(x) = 42 + \frac{6(2 - x^2) - 6x(4x+1)}{\sqrt{4x + 1}}\\\\\Longleftrightarrow f'(x) = 42 + \frac{12 - 6x^2 -24x^2-6x}{\sqrt{4x + 1}}$

$\Longleftrightarrow \boxed{f'(x)= 42 + \frac{6\left(2 - x - 5x^2 \right)}{\sqrt{4x + 1}}}$

Determinemos o domínio de $f':$

$4x + 1 > 0 \Longleftrightarrow 4x > -1\Longleftrightarrow x > -\frac{1}{4}.\\\\D_{f'} = \left\{x \in \mathbb{R}\, |\, x > -\frac{1}{4} \right\}.$

Calculemos o valor de f' para x = 2:

$\Longleftrightarrow f'(2) = 42 + \frac{6\left(2 - 2 - 5\cdot 2^2 \right)}{\sqrt{4\cdot 2 + 1}}\\\\\Longleftrightarrow f'(2) = 42 + \frac{6 \cdot (-20)}{\sqrt{9}}\\\\\Longleftrightarrow f'(2) = 42 - \frac{120}{3}\\\\\Longleftrightarrow f'(2) = 42 -40\\\\\Longleftrightarrow \boxed{f'(2) = 2.}$