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Suppose a function like $f(x)=ax^3+bx^2+cx ; a,b,c\in R \text{ and }a\ne0 $ so

it has a root $x$ at which $f(x)=0$ but at that point $f'(x)=c$ and we all know that it is the slope of the tangent i.e. $\tan \theta =c\implies \theta =\tan^{-1}c$. So will this tangent pass through origin?

If i talk in general then any function $g(x)$, at $x=0$, has a tangent that will pass through the origin?

anni saini
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  • You can see that the tangent to the function $f(x)=x^2+1$ at $x=0$ is a parallel to the x-axis that never pass through the origin. – Piquito Apr 20 '16 at 11:20
  • This is unclear. Firstly,$f'(0) = c$ is the gradient of the tangent through the origin. This is also $\tan \theta $, where $\theta $ is the angle subtended by this tangent and the x-axis. Secondly 'any function', g(x) is undefined. Very confusing. –  Apr 20 '16 at 11:21
  • The question stipulates it is a cubic since $a \ne 0$ –  Apr 20 '16 at 11:23
  • If $f(0)=0$ then it is trivial that any tangent to the curve at $x=0$ will pass through $(0,f(0))$, i.e. through the origin. If $f(0)\neq 0$, then plainly the tangent at $x=0$ cannot pass through the origin unless the slope is infinite. – hardmath Apr 20 '16 at 14:30

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The simple answer is that if the derivative exists at the point $x=0$ for a function $f:\mathbb{R}\to \mathbb{R}$and $f(0)=0$ then yes the tangent line to the curve at $x=0$ does pass through the origin.

To prove this note that the slope of the the tangent line is $f'(0)$ and the general formula for the tangent line is $y=f'(0)x+c$. We can find the constant $c$ because we know that at $x=0$ we have the point $(0,f(0))$. So substitution leads to $f(0)=0+c$ so $c=0$ iff $f(0)=0$. Obviously a straight line only passes through the origin if $c=0$.

The above argument shows that in fact this is the only condition under which a tangent line to a differentiable function at $x=0$ can pass through the origin.

K.Power
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  • I disagree that this is only case at which a function can have a tangent at $x=0$ passing through the origin. Consider $y = 1 + \sqrt[3]{x}$ on $\mathbb{R}$. The tangent at $x=0$ is vertical (infinite slope). – hardmath Apr 20 '16 at 18:33
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    @hardmath That depends on how you define tangent line. The derivative does not exist at $x=0$ so it's hard to be rigorous about it. But to note your objection I have put in the caveat of differentiability. – K.Power Apr 20 '16 at 18:59