Question

$find \: the \: derivative \: of \: the \: function \: f(x) = 2x ^{2} + 3x - 5 \: at \: x = - 1. \: also \: prove \: that \: \frac{d(f(x))}{dx \: } \: (0) + 3 \times \frac{d(f(x))}{dx} \: ( - 1) = 0.$
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Answer 1

Answer:

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

By Differnatating

$= 2. \frac{d(x {}^{2}) }{dx} + 3. \frac{d(x)}{dx} - 0(costant)$

$= 4x + 3$

F(x) = - 1

$= 4( - 1) + 3$

$= - 4 + 3$

$= - 1$

When f(x) = 0

$4(0) + 3 = 3$

Now add up as the condition is given,

F(0) + 3 F(-1)

$3 + 3( - 1)$

$= 3 - 3$

$= 0$

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