My Problem: What is the slope of the line normal to the graph of $f(x) = e^x-x^e-e$ at the point where the graph crosses the $x$-axis?
- a. $-0.288$
- b. $-0.110$
- c. $3.471$
- d. $9.106$
I found that $f(x)$ crosses the $x$ axis at $3.471$, but is that the answer? Do I have to plug it back into the equation? Thank you!
Recall: the equation of the line at $(x^*,y^*)$ for a function $f$ can be given by the point-slope form:
$$y - y^* = f'(x^*) \cdot (x - x^*)$$
(analogous to the $y=mx+b$ form for linear equations). The point where $f$ crosses the $x$ axis is a point where $y^* = 0$, and you found $x^* \approx 3.471$. All that remains for this part is to find $f'$ and then plug in everything as necessary above.
Shift it into the $y = mx+b$ form, then, where $m$ will be the slope of the tangent line at that point.
From there, recall that the "normal" line is just a line perpendicular to the given line. Suppose we have two lines,
$$y_1 = m_1 x + b_1 \;\;\;\;\; y_2 = m_2 x + b_2$$
If $y_2$ is perpendicular to $y_1$, then
$$m_2 = - \frac{1}{m_1}$$
i.e. the slopes of perpendicular lines are negative reciprocals of each other.