Calculus Derivatives - Finding the slope of a function

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We firstly determine the point where the graph crosses the x-axis. So we solve $f(x) = 0$ that is $$2x^{(2/3+3/3)}-5x^{(2/3)}=0\Rightarrow2x\cdot x^{(2/3)} - 5x^{(2/3)} = 0 \Leftrightarrow x^{(2/3)}(2x-5)=0$$ which gives $x = 5/2 \text{ or } x=0.$ Then we calculate the derivative of $y$ $$y'(x) = \frac{10}{3}x^{(2/3)} - \frac{10}{3}x^{(-1/3)}$$ which gives $$y'(x) = \frac{10}{3}\left(1-\frac{1}{x}\right)x^{(2/3)}$$ For $x=0$ the derivative does not exist so for $x = \frac{5}{2}$ we have that $$y'\left(\frac{5}{2}\right) = \frac{10}{3}\cdot\left(1-\frac{2}{5}\right)\cdot\left(\frac{5}{2}\right)^{(2/3)} =3.684$$.

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Updated on August 01, 2022

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  • user481710
    user481710 over 1 year

    I am having trouble solving this question:

    Consider the function $f(x)= 2x^{5/3} - 5x^{2/3}$

    Determine the slope of the tangent at the point where the graph crosses the x-axis.

  • Jose Antonio
    Jose Antonio over 9 years
    Please use Latex in the answers.