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I have been trying to understand the relationship between derivatives of functions, their tangents, and secant lines. I came across this thread, but it has not really helped me understand why certain functions, when derived, give a secant line. For example, if I have a function $$f(x) = x^3 + 2x^2 + 3$$ and its derivative, $$f'(x) = 3x^2 + 4x$$ when I graph them, the derivative is a second-degree function, so it will be a parabola and intersect the original function twice. From my understanding, this is a secant line, since it intersects the function two times, whereas a tangent only intersects once at a certain point. Since this is the derivative, how does it give me a secant line? It may be false that the tangent line can only intersect at one point, but that still does not make sense to me. If that is true, how would you know the difference between a tangent line and the secant line?

In my case here, how would I find the tangent line? it seems to me that functions with a degree greater than two will have this complication as well.

$$f(x) = x^n, n > 2$$

So, if someone would help me understand that relationship between these three concepts and why the tangent line may, in most cases, intersect the function more than once it would help me very much.

Thank you.

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The purpose of $f'(x)$, the derivative, is to give you the slope of a tangent. Meaning if you want to find the slope of the tangent line to $f(x)$ at $(x,y)$ you would plug in $x$ into $f'(x)$. The intersection of $f'(x)$ and $f(x)$ twice just means there are two values of $x$ where $f'(x)$ and $f(x)$ happen to be equal.

So basically what I'm saying is that the function $f'(x)$ is not the tangent line: there are an infinite number of tangent lines you can draw. You use $f'(x)$ simply to find the slope of the tangent line to a graph at a point.

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Recall that the value of the derivative function at a point $x=x_0$, that is $y_0=f'(x_0)$, represents the value of the slope for the tangent line to $f(x)$ at the point $(x_0, f(x_0))$.

Note that, as a particular case, the function $f(x)=ke^x$ is such that the slope at any point coincides with the value of the function at that point indeed $f'(x)=ke^x=f(x)$.

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