Find the derivative of $f (x) = 2 x^{2} + 3 x - 5$ at $x = - 1$ Also prove that $f^{'} (0) + 3 f^{'} (- 1) = 0$

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Solution

$f (x) = 2 x^{2} + 3 x - 5$
$f^{'} (x) = 4 x + 3$
$f^{'} (- 1) = 4 (- 1) + 3 = - 1$
$f^{'} 1 (0) = 4 (0) + 3 = 3$
$f^{'} 1 (0) + 3 f^{'} 1 (- 1) = 3 - 3 = 0$

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Similar questions

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

x = - 1

f^{'} (0) + 3 f^{'} (- 1) = 0

f (x) = 2^{x} + 3 x - 5 a t x = - 1

f^{^{'}} (0) + 3 f^{^{'}} (- 1) = 0

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

f^{'} (0) + 3 f^{'} (- 1)

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x (e^{1 / x} - 1)}{e^{1 / x} + 1} & x \neq 0 0 & x = 0 \end{matrix}

f^{^{'}} (0^{+})

f

x = 0

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