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Question

# Evaluate the function at the indicated values.$f ( x ) = x ^ { 3 } + 2 x ; \quad f ( - 2 ) , f ( - 1 ) , f ( 0 ) , f \left( \frac { 1 } { 2 } \right)$

Solution

Verified
Step 1
1 of 2

Since $f(x) = x^3 + 2x$,

$f(-2) = (-2)^3 + 2(-2) = -8 - 4 = -12$,

$f(1) = (1)^3 + 2(1) = 1 +2 =3$,

$f(0) = (0)^3 + 2(0) = 0 +0 =0$,

$f(\frac{1}{3}) = (\frac{1}{3})^3 + 2(\frac{1}{3}) = \frac{1}{27} +\frac{2}{3} =\frac{1}{27} +\frac{18}{27} = \frac{19}{27}$, and

$f(0.2) = (0.2)^3 + 2(0.2) = 0.008 +0.4 =0.408$.

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