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I want to apologise in advance for not having this in latex or some sort of neat code, I would be more than happy to learn how though.

Anyway, for the function $y=4(x-1)^{2/5}$ I see there appears to be a vertical tangent at $x=1$, but how can I know for certain the vertical tangent exists at $x=1$? Would I just solve for $f'(x)$, letting $x=1$? But what would that tell me?

Thanks.

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3 Answers 3

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Since $f'(1)$ is undefined, you need to consider $$\lim_{x \rightarrow 1} f'(x).$$ You will find that we have $$\lim_{x \rightarrow 1^{+}} f'(x)=\infty$$ and $$\lim_{x \rightarrow 1^{-}} f'(x)=-\infty.$$ In such a situation, where $f(1)$ is defined, we say that the curve $y=f(x)$ has a vertical tangent line at $x=1$.

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The slope of the tangent at $(x_0,y_0)$ is $f\prime(x_0)$ and hence the equation is $y-y_0=f\prime(x_0)(x-x_0)$

You can do the calculations now I believe.

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Yes, you would check if $f'(1)$ tends to $+\infty$ or $-\infty$

$$ \frac{d}{dx}4(x-1)^{2/5}=\frac{8}{5}(x-1)^{-3/5}\\ $$

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