Does the graph of the function $f$ have tangent line at the given points? If yes, what is the tangent line?
$f(x)=(x+2)^{3/5}$ at $x=-2$
solution: yes, $x=-2$
The derivative I found:
$$f'(x)=\frac 3{5(x+2)^\frac 25}$$
and then I get $f'(-2)=\frac 3{5\cdot 0}=\frac 30$ which is undefined.
anyone know how to go about this problem? I got the slope or whatever to be undefined or something, any tips/solution appreciated!
Best Answer
To find the slope, evaluate the derivative of your function at that specific point:
$$f(x) = (x+2)^{3/5}$$ $$f'(x) = \frac{3}{5}*(x+2)^{-2/5}$$ $$f'(-2) = \frac{3}{5*0} = \text{undefined}$$
Therefore, your tangent line will be the form $$x =a$$ where a is a constant.
It will be a vertical tangent line as well.
A line with a slope of $0$ will have the form $$y = a$$ And it will be a horizontal tangent line.
Since at x = -2 your slope is undefined, a = -2.
Therefore,
$$x = -2$$ is the tangent line.