Define $f:\R\to[0,\infty)$ by $f(x)$ $=$ $\sqrt{\x2 + 1}.$

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[1] Give the domain of $f.$

$\R$

$[0,\infty)$

[1] Give the range of $f.$

$[1,\infty)$

Define $f:\R\to\R$ by $f(x)$ $=$ $\sqrt[5]{x} - 7.$

Define $f:\R\to\R$ by $f(x)$ $=$ $\sqrt[7]{x} - 5.$

[1] Prove or reasonably explain that $f$ is 1-1.

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$f\left(x_{{}_{1}}\right)$ $=$ $f\left(x_{{}_{2}}\right)$

$\sqrt[5]{x_{{}_{1}} - 7}$ $=$ $\sqrt[5]{x_{{}_{2}} - 7}$

$\sqrt[5]{x_{{}_{1}}}$ $=$ $\sqrt[5]{x_{{}_{2}}}$

$\left(\sqrt[5]{x_{{}_{1}}}\right)^{5}$ $=$ $\left(\sqrt[5]{x_{{}_{2}}}\right)^{5}$

$x_{{}_{1}}$ $=$ $x_{{}_{2}}$

$f\left(x_{{}_{1}}\right)$ $=$ $f\left(x_{{}_{2}}\right)$

$\sqrt[7]{x_{{}_{1}} - 5}$ $=$ $\sqrt[7]{x_{{}_{2}} - 5}$

$\sqrt[7]{x_{{}_{1}}}$ $=$ $\sqrt[7]{x_{{}_{2}}}$

$\left(\sqrt[7]{x_{{}_{1}}}\right)^{7}$ $=$ $\left(\sqrt[7]{x_{{}_{2}}}\right)^{7}$

$x_{{}_{1}}$ $=$ $x_{{}_{2}}$

[1] Prove or reasonably explain that $f$ is onto.

Since $f$ is continuous, $\limv x,\tm\infty;f(x)$ $=$ $\tm\infty,$ and $\limv x,\infty;f(x)$ $=$ $\infty,$ $f$ is onto by the .

[1] Find the inverse of $f.$

[2]()

$y$ $=$ $\sqrt[5]{x} - 7$

$\sqrt[5]{x}$ $=$ $y + 7$

$x$ $=$ $\left(y + 7\right)^{5}$

$y$ $=$ $\sqrt[7]{x} - 5$

$\sqrt[7]{x}$ $=$ $y + 5$

$x$ $=$ $\left(y + 5\right)^{7}$

Define $f:\R\to\R$ by $f(x)$ $=$ $\x3 + 7x + 14.$

[1] Prove or reasonably explain that $f$ is 1-1.

Since $f'(x)$ $=$ $3\x2 + 7 \geq 7 > 0,$ $f$ is strictly increasing which implies that it is 1-1.

[2]()

[1] Calculate $\inv f;(\tm8)$

$\inv f;(\tm8)$ $=$ $\tm2$

[1] Calculate $[\inv f;]'(\tm8)$

$[\inv f;]'(\tm8)$ $=$ $\fric<1,f'(\tm2)>$ $=$ $\fric<1,19>$

Calculate each of the following.

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[1] $\invsec 2$ $=$ $\fric<\pi,3>$

$\invcsc 2$ $=$ $\fric<\pi,6>$

[1] $\sin\left(\invtan\fric<1,6>\right)$

Set $\theta$ $=$ $\invtan\fric<1,6>.$

$\tan\theta$ $=$ $\fric<1,6>$

$\sin\theta$ $=$ $\fric<1,\sqrt{37}>$

$\sin\left(\invtan\fric<1,6>\right)$ $=$ $\fric<1,\sqrt{37}>$

[1] Give the following as an algebraic expression in simplified form.

$\tan\left(\invsec w\right)$

$\sec\left(\invtan w\right)$

See class notes.

Differentiate.

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[1] $f(x)$ $=$ $\invsec x$

$f'(x)$ $=$ $\dfric<1,x\sqrt{\x2 - 1}>$

$g'(x)$ $=$ $\dfric<2x + 1,\sqrt{1 - (\x2 + x + 1)}>$

[1] $h(x)$ $=$ $\sqrt{\invtan x}$

$h'(x)$ $=$ $\dfric<1,2\sqrt{\invtan x}(\x2 + 1)>$

[1] Integrate.

$\dint\dfric<1,\sqrt{4 - \x2 }>dx$ $=$ $\invsin\left(\fric<x,2>\right) + C$

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