Tangent Lines & Linear Approximation

$f(2) = 2(4) - 6 + 1 = 3$, so the point is $(2, 3)$.

$f'(x) = 4x - 3$.

$m = f'(2) = 8 - 3 = 5$.

$y - 3 = 5(x - 2)$.

$y(0) = 5 \cdot 0 \cdot e^0 = 0$, so $(0, 0)$ is on the curve.

Product rule: $y' = 5e^{x} + 5xe^{x} = 5e^{x}(1 + x)$.

$m = y'(0) = 5 \cdot 1 \cdot 1 = 5$.

$y - 0 = 5(x - 0)$.

$y(1) = \sqrt{9} = 3$, so the point is $(1, 3)$.

$y = (1 + 8x)^{1/2}$, so $y' = \tfrac{1}{2}(1 + 8x)^{-1/2} \cdot 8 = \dfrac{4}{\sqrt{1 + 8x}}$.

$m = \dfrac{4}{\sqrt{9}} = \dfrac{4}{3}$.

$y - 3 = \dfrac{4}{3}(x - 1)$.

The point is given: $(5, 7)$.

Nothing to differentiate — the derivative's value is given.

$m = f'(5) = -2$.

$y - 7 = -2(x - 5)$.

The line $y = x - 4$ has slope $1$, so we need $m = 1$.

Quotient rule: $y' = \dfrac{(1)(x + 2) - (x - 2)(1)}{(x + 2)^2} = \dfrac{4}{(x + 2)^2}$.

$\dfrac{4}{(x + 2)^2} = 1 \Rightarrow (x + 2)^2 = 4 \Rightarrow x + 2 = \pm 2 \Rightarrow x = 0$ or $x = -4$.

$y(0) = \dfrac{-2}{2} = -1$ and $y(-4) = \dfrac{-6}{-2} = 3$. Points $(0, -1)$ and $(-4, 3)$.

$y + 1 = 1(x - 0)$ and $y - 3 = 1(x + 4)$.

Horizontal means slope $m = 0$.

$f'(x) = 3x^2 - 6x - 9$.

$3x^2 - 6x - 9 = 0 \Rightarrow 3(x - 3)(x + 1) = 0 \Rightarrow x = 3$ or $x = -1$.

$f(3) = 27 - 27 - 27 + 2 = -25$ and $f(-1) = -1 - 3 + 9 + 2 = 7$.

$y + 25 = 0(x - 3)$ and $y - 7 = 0(x + 1)$, that is $y = -25$ and $y = 7$.

$x + 4y = 1 \Rightarrow y = -\tfrac{1}{4}x + \tfrac{1}{4}$, slope $-\tfrac{1}{4}$. Perpendicular slope: $-1 \big/ \left(-\tfrac{1}{4}\right) = 4$.

$y' = 2x$.

$2x = 4 \Rightarrow x = 2$.

$y(2) = 4$, so the point is $(2, 4)$.

$y - 4 = 4(x - 2)$.

$f(1) = 3 - 1 = 2$, $f'(1) = 3$; $g(1) = -2 + 5 = 3$, $g'(1) = -2$.

$h(1) = f(1)g(1) = 2 \cdot 3 = 6$.

$h'(x) = f'(x)g(x) + f(x)g'(x)$.

$m = h'(1) = 3 \cdot 3 + 2 \cdot (-2) = 5$.

$y - 6 = 5(x - 1)$.

Already decoded: $f(1) = 2$, $f'(1) = 3$, $g(1) = 3$, $g'(1) = -2$.

$h(1) = \dfrac{2}{3}$.

$h'(x) = \dfrac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$.

$m = \dfrac{3 \cdot 3 - 2 \cdot (-2)}{3^2} = \dfrac{13}{9}$.

$y - \dfrac{2}{3} = \dfrac{13}{9}(x - 1)$.

$g(2) = -8 + 9 = 1$ and $g'(2) = -4$.

$h(2) = (4 + 1) \cdot 1 = 5$.

$h'(x) = 2x\,g(x) + (x^2 + 1)\,g'(x)$.

$m = h'(2) = 4 \cdot 1 + 5 \cdot (-4) = -16$.

$y - 5 = -16(x - 2)$.

Touching point $(a, a^2)$.

Slope from the derivative: $2a$. Slope from the points: $\dfrac{a^2 - (-3)}{a - 1} = \dfrac{a^2 + 3}{a - 1}$.

$\dfrac{a^2 + 3}{a - 1} = 2a \Rightarrow a^2 + 3 = 2a^2 - 2a \Rightarrow a^2 - 2a - 3 = 0 \Rightarrow a = 3$ or $a = -1$.

$a = 3$: point $(3, 9)$, slope $6$, so $y - 9 = 6(x - 3)$. $a = -1$: point $(-1, 1)$, slope $-2$, so $y - 1 = -2(x + 1)$. Both pass through $(1, -3)$.

Touching point $(a, e^{a})$.

From the derivative: $e^{a}$. From the points: $\dfrac{e^{a} - 0}{a - 0} = \dfrac{e^{a}}{a}$.

$\dfrac{e^{a}}{a} = e^{a} \Rightarrow a = 1$ (since $e^{a} \neq 0$).

Point $(1, e)$, slope $e$: $y - e = e(x - 1)$, which simplifies to $y = ex$ and passes through $(0, 0)$.

$f(x) = \sqrt{x}$, $a = 4$ (a perfect square close to $4.1$).

$f(4) = 2$.

$f'(x) = \dfrac{1}{2\sqrt{x}}$, so $f'(4) = \dfrac{1}{4}$.

$L(x) = 2 + \dfrac{1}{4}(x - 4)$.

$L(4.1) = 2 + \dfrac{1}{4}(0.1) = 2.025$.

$f(x) = x^{1/3}$, $a = 8$.

$f(8) = 2$.

$f'(x) = \tfrac{1}{3}x^{-2/3}$, so $f'(8) = \tfrac{1}{3} \cdot \tfrac{1}{4} = \tfrac{1}{12}$.

$L(x) = 2 + \dfrac{1}{12}(x - 8)$.

$L(8.3) = 2 + \dfrac{0.3}{12} = 2.025$.

$a = 3$ — given, and $3.2$ is close to it.

$f(3) = -1$.

$f'(3) = 2$.

$L(x) = -1 + 2(x - 3)$.

$L(3.2) = -1 + 2(0.2) = -0.6$.

$f(x) = \sin x$ and $g(x) = \cos x$, both at $a = 0$; $0.05$ is close to $0$ (radians).

$\sin 0 = 0$ and $\cos 0 = 1$.

$(\sin x)' = \cos x \to 1$ at $0$; $(\cos x)' = -\sin x \to 0$ at $0$.

$L_{\sin}(x) = 0 + 1 \cdot x = x$ and $L_{\cos}(x) = 1 + 0 \cdot x = 1$.

$\sin(0.05) \approx 0.05$ and $\cos(0.05) \approx 1$.