# Thread: Finding exact zeros and tangent lines.

1. ## Finding exact zeros and tangent lines.

So, here's the problem:

1) Let f be the function defined by $\displaystyle f(x) = -A + ln(x^2)$, where A is the number of letters in your last name. (For me, that would be 6).
(a) Find the exact zeros of f (using algebra, not your calculator).
(b) Write an equation for the tangent line to the graph of f at x = -1.

Can someone explain to me how to get started solving these questions?

2. To begin with, $\displaystyle x$ is a zero of $\displaystyle f(x)$ when

$\displaystyle -6 + \ln x^2=0.$

Solving this equation gives us the zeros of $\displaystyle f(x)$. To find the equation of the tangent line at $\displaystyle x=-1$, we differentiate $\displaystyle f(x)$ to find its slope at $\displaystyle x=-1$, remembering that the equation for a line is

$\displaystyle y-y_1=m(x-x_1).$

3. Originally Posted by Rumor
So, here's the problem:

1) Let f be the function defined by $\displaystyle f(x) = -A + ln(x^2)$, where A is the number of letters in your last name. (For me, that would be 6).
(a) Find the exact zeros of f (using algebra, not your calculator).
(b) Write an equation for the tangent line to the graph of f at x = -1.

Can someone explain to me how to get started solving these questions?

for a)

$\displaystyle -A + \ln (x^2) =0$

$\displaystyle A = \ln (x^2)$

$\displaystyle e^{A} = x^2$

$\displaystyle x = \mp e^{\frac{A}{2}}$

for 2)

$\displaystyle f(x) = -A + \ln (x^2)$

$\displaystyle f'(x) = 0 + \frac{2x}{x^2}$

$\displaystyle f'(-1) = \frac{-2}{1} = -2$ the slope of the function at x=-1

the equation of the tangent line when x=-1 , first find y when x=-1

$\displaystyle f(-1) = -A + \ln (x^2) = -A + 0 =-A$

the equation

$\displaystyle y-(-A ) = slope (x-(-1))$

$\displaystyle y+A = -2(x+1)$

4. Thank you to the both of you! That was really helpful.