[Math] Vertical cusp vertical tangent question

calculusderivatives

Can anyone help me solve the following two problems:

  • Does the following function have a vertical tangent or vertical cusp at $c=0$

$$f(x)=(3+x^{\frac{2}{5}})$$

I got for the derivative

$$f'(x)=\frac{2}{5\sqrt[5]{x}^3}$$

Now I think this would be a cusp because as x approaches zero from the left and right f(x) approches infinity.

  • My second question is with the function $f(x)=|{(x+8)^{1/3}}|$.

For my derivative I got

$$f'(x)=\frac{1}{3}(x+8)^{\frac{-2}{3}},\;x > -8$$

$$f'(x)=\frac{-1}{3}(x+8)^{\frac{-2}{3}},\;x < -8$$

Would this not be a tangent because of the square expoent.

Best Answer

Yes, your first function has a cusp at $x = 0$, because the derivative $f'(x)$ approaches $\infty$ as $x \to 0^+$ and as $x\to 0^-$. We can see this at the point $(0, 3)$:

$\quad f(x)=\left(3+x^{\large\frac{2}{5}}\right)$ (see blue curve)

enter image description here

Your second function also has a cusp: at $x = -8$

Can you try to rewrite your derivative, or redefine if you mean "for $x \geq 0, f'(x) = ...$ and for $x < 0, f'(x) = ...?$.

$\quad f(x)=|(x+8)^{1/3}|$

enter image description here

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