Find f(g(x)) f(x)=2x^2-1 , g(x) = square root of (x+1)/2

$f (x) = 2 x^{2} - 1$ , $g (x) = \sqrt{\frac{x + 1}{2}}$

Set up the composite result function.

$f (g (x))$

Evaluate $f (\sqrt{\frac{x + 1}{2}})$ by substituting in the value of $g$ into $f$ .

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

Simplify each term.

Rewrite $\sqrt{\frac{x + 1}{2}}$ as $\frac{\sqrt{x + 1}}{\sqrt{2}}$ .

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

Multiply $\frac{\sqrt{x + 1}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt{x + 1}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Combine using the product rule for radicals.

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

Apply the product rule to $\frac{\sqrt{(x + 1) \cdot 2}}{2}$ .

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

Simplify the numerator.

Rewrite ${\sqrt{(x + 1) \cdot 2}}^{2}$ as $(x + 1) \cdot 2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(x + 1) \cdot 2}$ as ${((x + 1) \cdot 2)}^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Move $2$ to the left of $x$ .

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

Multiply $2$ by $1$ .

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

Factor $2$ out of $2 x + 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (x) + 2 (1)$ .

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

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

$f (\sqrt{\frac{x + 1}{2}}) = 2 (\frac{2 (x + 1)}{4}) - 1$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$f (\sqrt{\frac{x + 1}{2}}) = 2 (\frac{2 (x + 1)}{2 (2)}) - 1$

Cancel the common factor.

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

Rewrite the expression.

$f (\sqrt{\frac{x + 1}{2}}) = \frac{2 (x + 1)}{2} - 1$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\sqrt{\frac{x + 1}{2}}) = \frac{2 (x + 1)}{2} - 1$

Divide $x + 1$ by $1$ .

$f (\sqrt{\frac{x + 1}{2}}) = x + 1 - 1$

Combine the opposite terms in $x + 1 - 1$ .

Subtract $1$ from $1$ .

$f (\sqrt{\frac{x + 1}{2}}) = x + 0$

Add $x$ and $0$ .

$f (\sqrt{\frac{x + 1}{2}}) = x$

Find f(g(x)) f(x)=2x^2-1 , g(x) = square root of (x+1)/2

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