I am wondering if the slope of a point on a straight line is equal to the overall slope of that straight line, Because I've just started with calculus, and we took an intro to differentials and derivatives And we say that the slope of the tangent line is equal to the slope of the curve at that given point. but is the overall slope of the tangent line equal to the slope of the curve at the given point, or is the slope of that point on the tangent line and only that point is equal to the slope of the curve at the given point. from here came my question since the tangent line is a straight line, is the slope of a point that is on a straight line the same as the overall slope of that line ( which is straight). Because here I'm not talking about the slope between two points on a straight line which is definitely equal to the over all slope of that line, but I'm talking about the slope at only one point. This question may seem stupid to some but answer me according to my stupidity. Also if I'm talking about the slope at only one point isn't that instant change?
is the slope of a point on a straight line the same as the overall slope of that straight line
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The slope $m$ of the tangent line between two points $a,b$ with $a<b$ on a function $f$ is given by $$ m=\frac{f(b)-f(a)}{b-a}. $$ This is the standard "rise over run" definition we are familiar with.
The slope $m$ at a point on $f$ simply means taking the slope of the tangent line of $a$ and $b$ as they get closer and closer to each other. If we define $b=a+h$ and we consider what happens when $h$ gets infinitely small, we have the definition of the derivative of $f$ at $a$ expressed with the idea of a limit: $$ m=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}. $$ Since $f$ is linear (it has the property of linearity), we can rewrite the limit as $$ m=\lim_{h\to 0}\frac{f(a)+f(h)-f(a)}{h}=\lim_{h\to 0}\frac{f(h)}{h}. $$ We see now that the slope $m$ does not depend on $a$ anymore. In other words, the slope will be the same anywhere we look at the line.
Side note: the final equation is only possible if $f$ is linear, not affine. However, transforming an affine function so that it is fixed to the origin (becomes linear) does not affect its derivative.
Does that answer your question? Please let me know if I've explained anything unclearly or incorrectly!
Watch out in your definition: the slope is referred to a line, and not "of a point" as you write. What the definition of derivative tells you is the following. Take a function $f$. For every point $x$ in the domain of $f$, if you consider the tangent line at $f(x)$, that tangent line has a certain slope, which is the derivative of the function evaluated at that point $x$. When you change the point $x$, the tangent line at the new $f(x)$ has a different slope (in general).
If you want, the derivative $f^\prime$ is also a function of $x$, because by plugging different points $x$ into $f^\prime$, you obtain different values $f^\prime (x)$, in general.