Find a first-degree polynomial function P1 whose value and slope agree with the value and slope of $f$ at $x = c$. What is $P _ { 1 }$ called?
$$f ( x ) = \tan x , c = - \frac { \pi } { 6 }$$
A)
$3 y + \sqrt { 3 } = - 4 \left( x + \frac { \pi } { 6 } \right) ;$ tangent line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
B)
$3 y + \sqrt { 3 } = 4 \left( x + \frac { \pi } { 6 } \right) ;$ secant line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
C)
$3 y + \sqrt { 2 } = 4 \left( x + \frac { \pi } { 6 } \right) ;$ tangent line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
D)
$3 x + \sqrt { 3 } = 4 \left( y + \frac { \pi } { 6 } \right) ;$ differential of $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
E)
$3 y + \sqrt { 3 } = 4 \left( x + \frac { \pi } { 6 } \right) ;$ tangent line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$

Question

Find a first-degree polynomial function P1 whose value and slope agree with the value and slope of $f$ at $x = c$. What is $P _ { 1 }$ called?
$$f ( x ) = \tan x , c = - \frac { \pi } { 6 }$$
A)
$3 y + \sqrt { 3 } = - 4 \left( x + \frac { \pi } { 6 } \right) ;$ tangent line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
B)
$3 y + \sqrt { 3 } = 4 \left( x + \frac { \pi } { 6 } \right) ;$ secant line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
C)
$3 y + \sqrt { 2 } = 4 \left( x + \frac { \pi } { 6 } \right) ;$ tangent line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
D) 
$3 x + \sqrt { 3 } = 4 \left( y + \frac { \pi } { 6 } \right) ;$ differential of $y = f ( x )$ at $x = - \frac { \pi } { 6 }$
E)
$3 y + \sqrt { 3 } = 4 \left( x + \frac { \pi } { 6 } \right) ;$ tangent line to $y = f ( x )$ at $x = - \frac { \pi } { 6 }$

Quizplus · Accepted Answer

Since we are asked to find a first-degree polynomial function that agrees with the value and slope of $f(x)$ at $x=c$, we want to find the equation of the tangent line to $f(x)$ at $x=c$, which will be in the form of a first-degree polynomial function. The formula for the equation of a tangent line at a point $(c,f(c))$ on a curve with equation $f(x)$ is:

$$y - f(c) = f'(c)(x-c)$$

where $f'(c)$ is the slope of the curve at the point $(c,f(c))$.

In this case, we have $f(x) = \tan x$ and $c = -\pi/6$, so we need to find $f'(-\pi/6)$ first. Using the derivative of the tangent function, we have:

$$f'(x) = \sec^2 x \quad \Rightarrow \quad f'(-\pi/6) = \sec^2(-\pi/6) = \frac{4}{3}$$

So, the equation of the tangent line to $f(x) = \tan x$ at $x=-\pi/6$ is:

$$y - \tan(-\pi/6) = \frac{4}{3}(x+\pi/6)$$

Simplifying:

$$3y + \sqrt{3} = 4\left(x + \frac{\pi}{6}\right)$$

Therefore, the best choice is (E).

Find a First-Degree Polynomial Function P1 Whose Value and Slope fff

Find a First-Degree Polynomial Function P1 Whose Value and Slope $f$