$f (x) = x^{3} - 3 x + 5$ at $x = 1.99$

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Solution

Consider the given function

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

On differentiation with respect to $x$

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

At $x = 1.99$

Then,

$f^{'} (1.99) = 3 {(1.99)}^{2} - 3 \times 1.99 + 5$

$f^{'} (1.99) = 10.9103 (a p p o r x .)$

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Similar questions

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

x = 1.99

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

x = 1.99

-

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

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

x = - 1

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