Evaluate the Function h(x)=5x^3-3x^2+2x+4 , x=-1/3

$h (x) = 5 x^{3} - 3 x^{2} + 2 x + 4$ , $x = - \frac{1}{3}$

Step 1

Replace the variable $x$ with $- \frac{1}{3}$ in the expression.

$h (- \frac{1}{3}) = 5 {(- \frac{1}{3})}^{3} - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Step 2

Simplify the result.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{3}$ .

$h (- \frac{1}{3}) = 5 ({(- 1)}^{3} {(\frac{1}{3})}^{3}) - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Apply the product rule to $\frac{1}{3}$ .

$h (- \frac{1}{3}) = 5 ({(- 1)}^{3} (\frac{1^{3}}{3^{3}})) - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Raise $- 1$ to the power of $3$ .

$h (- \frac{1}{3}) = 5 (- \frac{1^{3}}{3^{3}}) - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

One to any power is one.

$h (- \frac{1}{3}) = 5 (- \frac{1}{3^{3}}) - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Raise $3$ to the power of $3$ .

$h (- \frac{1}{3}) = 5 (- \frac{1}{27}) - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Multiply $5 (- \frac{1}{27})$ .

Multiply $- 1$ by $5$ .

$h (- \frac{1}{3}) = - 5 (\frac{1}{27}) - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Combine $- 5$ and $\frac{1}{27}$ .

$h (- \frac{1}{3}) = \frac{- 5}{27} - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Move the negative in front of the fraction.

$h (- \frac{1}{3}) = - \frac{5}{27} - 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 4$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{3}$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - 3 ({(- 1)}^{2} {(\frac{1}{3})}^{2}) + 2 (- \frac{1}{3}) + 4$

Apply the product rule to $\frac{1}{3}$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - 3 ({(- 1)}^{2} (\frac{1^{2}}{3^{2}})) + 2 (- \frac{1}{3}) + 4$

Raise $- 1$ to the power of $2$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - 3 (1 (\frac{1^{2}}{3^{2}})) + 2 (- \frac{1}{3}) + 4$

Multiply $\frac{1^{2}}{3^{2}}$ by $1$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - 3 \frac{1^{2}}{3^{2}} + 2 (- \frac{1}{3}) + 4$

One to any power is one.

$h (- \frac{1}{3}) = - \frac{5}{27} - 3 \frac{1}{3^{2}} + 2 (- \frac{1}{3}) + 4$

Raise $3$ to the power of $2$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - 3 (\frac{1}{9}) + 2 (- \frac{1}{3}) + 4$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

$h (- \frac{1}{3}) = - \frac{5}{27} + 3 (- 1) (\frac{1}{9}) + 2 (- \frac{1}{3}) + 4$

Factor $3$ out of $9$ .

$h (- \frac{1}{3}) = - \frac{5}{27} + 3 \cdot (- 1 \frac{1}{3 \cdot 3}) + 2 (- \frac{1}{3}) + 4$

Cancel the common factor.

$h (- \frac{1}{3}) = - \frac{5}{27} + 3 \cdot (- 1 \frac{1}{3 \cdot 3}) + 2 (- \frac{1}{3}) + 4$

Rewrite the expression.

$h (- \frac{1}{3}) = - \frac{5}{27} - 1 (\frac{1}{3}) + 2 (- \frac{1}{3}) + 4$

Rewrite $- 1 (\frac{1}{3})$ as $- (\frac{1}{3})$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - (\frac{1}{3}) + 2 (- \frac{1}{3}) + 4$

Multiply $2 (- \frac{1}{3})$ .

Multiply $- 1$ by $2$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - \frac{1}{3} - 2 (\frac{1}{3}) + 4$

Combine $- 2$ and $\frac{1}{3}$ .

$h (- \frac{1}{3}) = - \frac{5}{27} - \frac{1}{3} + \frac{- 2}{3} + 4$

Move the negative in front of the fraction.

$h (- \frac{1}{3}) = - \frac{5}{27} - \frac{1}{3} - \frac{2}{3} + 4$

Combine fractions.

Combine the numerators over the common denominator.

$h (- \frac{1}{3}) = 4 + \frac{- 5}{27} + \frac{- 1 - 2}{3}$

Subtract $2$ from $- 1$ .

$h (- \frac{1}{3}) = 4 + \frac{- 5}{27} + \frac{- 3}{3}$

Find the common denominator.

Write $4$ as a fraction with denominator $1$ .

$h (- \frac{1}{3}) = \frac{4}{1} + \frac{- 5}{27} + \frac{- 3}{3}$

Multiply $\frac{4}{1}$ by $\frac{27}{27}$ .

$h (- \frac{1}{3}) = \frac{4}{1} \cdot \frac{27}{27} + \frac{- 5}{27} + \frac{- 3}{3}$

Multiply $\frac{4}{1}$ by $\frac{27}{27}$ .

$h (- \frac{1}{3}) = \frac{4 \cdot 27}{27} + \frac{- 5}{27} + \frac{- 3}{3}$

Multiply $\frac{- 3}{3}$ by $\frac{9}{9}$ .

$h (- \frac{1}{3}) = \frac{4 \cdot 27}{27} + \frac{- 5}{27} + \frac{- 3}{3} \cdot \frac{9}{9}$

Multiply $\frac{- 3}{3}$ by $\frac{9}{9}$ .

$h (- \frac{1}{3}) = \frac{4 \cdot 27}{27} + \frac{- 5}{27} + \frac{- 3 \cdot 9}{3 \cdot 9}$

Multiply $3$ by $9$ .

$h (- \frac{1}{3}) = \frac{4 \cdot 27}{27} + \frac{- 5}{27} + \frac{- 3 \cdot 9}{27}$

Combine the numerators over the common denominator.

$h (- \frac{1}{3}) = \frac{4 \cdot 27 - 5 - 3 \cdot 9}{27}$

Simplify each term.

Multiply $4$ by $27$ .

$h (- \frac{1}{3}) = \frac{108 - 5 - 3 \cdot 9}{27}$

Multiply $- 3$ by $9$ .

$h (- \frac{1}{3}) = \frac{108 - 5 - 27}{27}$

Simplify by subtracting numbers.

Subtract $5$ from $108$ .

$h (- \frac{1}{3}) = \frac{103 - 27}{27}$

Subtract $27$ from $103$ .

$h (- \frac{1}{3}) = \frac{76}{27}$

The final answer is $\frac{76}{27}$ .

$\frac{76}{27}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$\frac{76}{27}$

Decimal Form:

$2.\overline{814}$

Mixed Number Form:

$2 \frac{22}{27}$

Step 4

Evaluate the Function h(x)=5x^3-3x^2+2x+4 , x=-1/3

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