If $f (x) = 2 x^{2} + 3 x - 5$ , then what is $f^{'} (0) + 3 f^{'} (- 1)$ equal to?

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Solution

The correct option is A $0$
$f (x) = 2 x^{2} + 3 x - 5$
$\Rightarrow f^{'} (x) = 4 x + 3$
$f^{'} (0) = 4 \times 0 + 3 = 3 f^{'} (- 1) = 4 (- 1) + 3 = - 1$
So, $f^{'} (0) + 3 f^{'} (- 1) = 3 + 3 (- 1) = 0$

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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^{2} + 3 x - 5

x = - 1

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

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{x - 1}{2 x^{2} - 7 x + 5}, x \neq 0 \frac{- 1}{3}, x = 1 \end{matrix}

f^{'} (1)

f (x) = \frac{x^{2}}{2}

0 \leq x \leq 1

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

1 \leq x \leq 2

f^{''}

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