Question

1. 0/1 points find an equation of the tangent line to the curve at the given point. ( y = sqrt{2 - 47x} ), ( (-1, 7) )

Explanation:

Step1: Rewrite function for differentiation

$y=(2-47x)^{\frac{1}{2}}$

Step2: Apply chain rule to find derivative

Let $u=2-47x$, so $y=u^{\frac{1}{2}}$.
$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2}u^{-\frac{1}{2}} \cdot (-47) = -\frac{47}{2(2-47x)^{\frac{1}{2}}}$

Step3: Calculate slope at $x=-1$

Substitute $x=-1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\bigg|_{x=-1}=-\frac{47}{2\sqrt{2-47(-1)}} = -\frac{47}{2\sqrt{49}} = -\frac{47}{14}$

Step4: Use point-slope form for tangent line

Point-slope formula: $y-y_1=m(x-x_1)$, where $(x_1,y_1)=(-1,7)$ and $m=-\frac{47}{14}$.
$y-7=-\frac{47}{14}(x+1)$

Step5: Simplify to slope-intercept form

$y=-\frac{47}{14}x - \frac{47}{14} + 7 = -\frac{47}{14}x + \frac{-47+98}{14} = -\frac{47}{14}x + \frac{51}{14}$

Answer:

$y = -\frac{47}{14}x + \frac{51}{14}$ (or in standard form: $47x + 14y = 51$)