Cube root function pls help me....

[minhthai2312]

y= expression^(1/3) (x)

because i don't know how to include the math equation in here.... so at the equation above mean y= cube root x
=> y= expression^(1/3) (x) = y= cube root (x)
a/ Find an equation of the tangent to the curve y= expression^(1/3) (x) at the point where x=27. Calculate the coordinates of the point where the tangent cuts the x-asix. Verify that it is a tangent by graphing both the cube root function and your equation in graphmatica
b/ let x(0) be any positive number and let Q be the point on the curve y= expression^(1/3) (x) where x=x(0). Find an expression for the point where the tangent at Q intersects with the x-axis / Conclude and make an illustration

Pls help me with that problem....

 

[Deleted member 4993]

y= expression^(1/3) (x)
a/ Find an equation of the tangent to the curve y= expression^(1/3) (x) at the point where x=27. Calculate the coordinates of the point where the tangent cuts the x-asix. Verify that it is a tangent by graphing both the cube root function and your equation in graphmatica
b/ let x(0) be any positive number and let Q be the point on the curve y= expression^(1/3) (x) where x=x(0). Find an expression for the point where the tangent at Q intersects with the x-axis / Conclude and make an illustration

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[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ x^{1/3}, \ f \ ' \ (x) \ = \ \frac{1}{3x^{2/3}}.\)

\(\displaystyle f \ ' \ (27) \ = \ \frac{1}{27} \ = \ m, \ f(27) \ = \ 3.\)

\(\displaystyle (27,3), \ m \ = \ \frac{1}{27}.\)

\(\displaystyle y-3 \ = \ \frac{1}{27}(x-27),\ y \ = \ \frac{x}{27}+2.\)

\(\displaystyle See \ graph.\)

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