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Linear Functions

Last updated

Jun 20, 2024
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- Linear Algebra and Partial Fractions Review
- Slope Field Review

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Definition: Linear Function

A function $f$ is linear if $f(a \mathbf{x}+b \mathbf{y})=a f(\mathbf{x})+b f(\mathbf{y})$ for scalars $a$ and $b$ .

Or equivalently $f$ is linear if (1.) $f(a \mathbf{x})=a f(\mathbf{x})$ and (2.) $f(\mathbf{x}+\mathbf{y})=f(\mathbf{x})+f(\mathbf{y})$

Theorem $\PageIndex{1}$

If $f$ is linear, then $f(\mathbf{0})=\mathbf{0}$ .

Proof

Observe that $f(\mathbf{0})=f(0 \cdot \mathbf{0})=0 \cdot f(\mathbf{0})=\mathbf{0}$

Example $\PageIndex{1}$

Determine if the following functions are linear or not.

1) $f: \mathbb{R} \to \mathbb{R}, f(x)=2x$

2) $f: \mathbb{R} \to \mathbb{R}, f(x)=2x+3$

Solution

The first function is linear as $f(a\mathbf{x}+b\mathbf{y}) = 2(a\mathbf{x}+b\mathbf{y}) = 2a\mathbf{x}+2b\mathbf{y} = af(\mathbf{x})+bf(\mathbf{y})$ .

The second function is not linear as we have the counter example of when $\mathbf{x}=0$ and $a=2$ observe that $f(2 \cdot 0) = 0+3 = 3$ but $2 \cdot f(0) = 2(0+3)=6$ and so $f(a\cdot x) \neq a\cdot f(x)$ .

Question

When is a line $f(x)=mx+b$ a linear function?

Example $\PageIndex{2}$

Determine if the following function is linear, $f: R^2 \rightarrow R^2 f\left(\left(x_1, x_2\right)\right)=\left(2 x_1, x_1+x_2\right)$ .

Solution

Let $\mathbf{x}=\left(x_1, x_2\right), \mathbf{y}=\left(y_1, y_2\right)$
$\notag \begin{array}{l} a \mathbf{x}+b \mathbf{y} \\ \left(a x_1+b y_1, a x_1, x_2\right)+b\left(y_1, y_2\right)=\left(a x_1, a x_2\right)+\left(b y_1, b y_2\right)= \\ \begin{array}{l} f\left(a x_1+b y_1, a x_2+b y_2\right) \\ =\left(2\left(a x_1+b y_1\right), a x_1+b y_1+a x_2+b y_2\right) \\ =\left(2 a x_1+2 b y_1, a x_1+a x_2+b y_1+b y_2\right) \\ =\left(2 a x_1, a x_1+a x_2\right)+\left(2 b y_1, b y_1+b y_2\right) \\ =a\left(2 x_1, x_1+x_2\right)+b\left(2 y_1, y_1+y_2\right) \\ =a f\left(\left(x_1, x_2\right)\right)+b f\left(\left(y_1, y_2\right)\right) \end{array} \end{array}$

Thus the function is linear.

Example $\PageIndex{3}$

Show that $D$ : set of all differential functions $\rightarrow$ set of all functions, $D(f)=f^{\prime}$ is a linear function

Solution

Observe, $D(a f+b g)=(a f+b g)^{\prime}=a f^{\prime}+b g^{\prime}=a D(f)+b D(g)$

Example $\PageIndex{4}$

Show that Given $a, b$ real numbers, $I$ : set of all integrable functions on $[\mathrm{a}, \mathrm{b}] \rightarrow R$ , $I(f)=\int_a^b f$ is a linear function.

Solution

Observe $I(s f+t g)=\int_a^b s f+t g=s \int_a^b f+t \int_a^b g=$ $s I(f)+t I(g)$ .

Example $\PageIndex{5}$

The inverse of a linear function is linear (when the inverse exists).

Solution

Suppose $f^{-1}(x)=c, f^{-1}(y)=d$ .
Then $f(c)=x$ and $f(d)=y$ and
$f(a c+b d)=a f(c)+b f(d)=a x+b y .$

Hence $f^{-1}(a x+b y)=a c+b d=a f^{-1}(x)+b f^{-1}(y)$ .

Example $\PageIndex{6}$

Example 6.) $D$ : set of all twice differential functions $\rightarrow$ set of all functions, $L(f)=a f^{\prime \prime}+b f^{\prime}+c f$

Solution

$\notag \begin{aligned} L(s f+t g) & =a(s f+t g)^{\prime \prime}+b(s f+t g)^{\prime}+c(s f+t g) \\ & =s a f^{\prime \prime}+t a g^{\prime \prime}+s b f^{\prime}+t b g^{\prime}+s c f+t c g \\ & =s\left(a f^{\prime \prime}+b f^{\prime}+c f\right)+t\left(a g^{\prime \prime}+b g^{\prime}+c g\right) \\ & =s L(f)+t L(g) \end{aligned}$

Corollary $\PageIndex{1}$

If $\psi_1, \psi_2$ are solutions to $a f^{\prime \prime}+b f^{\prime}+$ cf $=0$ , then $3 \psi_1+5 \psi_2$ is also a solution to $a f^{\prime \prime}+b f^{\prime}+c f=0$ .

Proof

Since $\psi_1, \psi_2$ are solutions to $a f^{\prime \prime}+b f^{\prime}+c f=0$ , $L\left(\psi_1\right)=0$ and $L\left(\psi_2\right)=0$ .

Hence
$\notag \begin{aligned} L\left(3 \psi_1+5 \psi_2\right) & =3 L\left(\psi_1\right)+5 L\left(\psi_2\right) \\ & =3(0)+5(0)=0 \end{aligned}$

Thus $3 \psi_1+5 \psi_2$ is also a solution to $a f^{\prime \prime}+b f^{\prime}+c f=0$

Corollary $\PageIndex{2}$

If $\psi_1$ is a solution to $a f^{\prime \prime}+b f^{\prime}+c f=h$ and $\psi_2$ is a solution to $a f^{\prime \prime}+b f^{\prime}+c f=k$ , then $3 \psi_1+5 \psi_2$ is a solution to $a f^{\prime \prime}+b f^{\prime}+c f=3 h+5 k$ .

Proof

Since $\psi_1$ is a solution to $a f^{\prime \prime}+b f^{\prime}+c f=h, L\left(\psi_1\right)=h$ .
Since $\psi_2$ is a solution to $a f^{\prime \prime}+b f^{\prime}+c f=k, L\left(\psi_2\right)=k$ .
Hence $L\left(3 \psi_1+5 \psi_2\right)=3 L\left(\psi_1\right)+5 L\left(\psi_2\right)$
$\notag =3 h+5 k .$

Thus $3 \psi_1+5 \psi_2$ is also a solution to
$\notag a f^{\prime \prime}+b f^{\prime}+c f=3 h+5 k$

Definition: Linear Function

Theorem 1\PageIndex{1}

Proof

Example 1\PageIndex{1}

Solution

Question

Example 2\PageIndex{2}

Solution

Example 3\PageIndex{3}

Solution

Example 4\PageIndex{4}

Solution

Example 5\PageIndex{5}

Solution

Example 6\PageIndex{6}

Solution

Corollary 1\PageIndex{1}

Proof

Corollary 2\PageIndex{2}

Proof

Support Center

How can we help?

Theorem $\PageIndex{1}$

Example $\PageIndex{1}$

Example $\PageIndex{2}$

Example $\PageIndex{3}$

Example $\PageIndex{4}$

Example $\PageIndex{5}$

Example $\PageIndex{6}$

Corollary $\PageIndex{1}$

Corollary $\PageIndex{2}$