Algebra Examples

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

Step 1

Set up the composite result function.

$h (g (x))$

Step 2

Evaluate $h (\frac{x}{4 - x})$ by substituting in the value of $g$ into $h$ .

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

Step 3

Simplify each term.

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Step 3.1

Apply the product rule to $\frac{x}{4 - x}$ .

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

Step 3.2

Apply the product rule to $\frac{x}{4 - x}$ .

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

Step 3.3

Combine $- 2$ and $\frac{x^{2}}{{(4 - x)}^{2}}$ .

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Simplify with commuting.

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Step 4.1

Reorder $4$ and $- x$ .

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

Step 4.2

Reorder $4$ and $- x$ .

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