Solve f(x)=1/3*(-2)^3+3*-1

$$f ( x ) = \frac { 1 } { 3 } \cdot ( - 2 ) ^ { 3 } + 3 \cdot - 1$$

Evaluate

$$-\frac{17}{3}\approx -5.666666667$$

Solution Steps

Calculate -2 to the power of 3 and get -8.

$$\frac{1}{3}\left(-8\right)+3\left(-1\right)$$

Multiply \frac{1}{3} and -8 to get \frac{-8}{3}.

$$\frac{-8}{3}+3\left(-1\right)$$

Fraction \frac{-8}{3} can be rewritten as -\frac{8}{3} by extracting the negative sign.

$$-\frac{8}{3}+3\left(-1\right)$$

Multiply 3 and -1 to get -3.

$$-\frac{8}{3}-3$$

Convert 3 to fraction \frac{9}{3}.

$$-\frac{8}{3}-\frac{9}{3}$$

Since -\frac{8}{3} and \frac{9}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-8-9}{3}$$

Subtract 9 from -8 to get -17.

$$-\frac{17}{3}$$

Factor

$$-\frac{17}{3} = -5\frac{2}{3} = -5.666666666666667$$