Thread: a question about tangent lines and slopes

the question is:
Find the points on the graph of y = x ^ (3/2) - x ^ (1/2) at which the tangent line is parallel to the line y - x = 3.

i have derived the first function to find the tangent line then i have derived it again to find it's slope, then derived the second function to get it's slope, then i have set the first slope = the second slope anyway this is what i got lastly:

(3/4) x ^ (-1/2) + (1/4) x ^ (-3/2) = 1
i could not get the value of x from this equation.is there any formula that will get me the value of x like the quadratic formula?

Originally Posted by mHadad

the question is:
Find the points on the graph of y = x ^ (3/2) - x ^ (1/2) at which the tangent line is parallel to the line y - x = 3.

i have derived the first function to find the tangent line then i have derived it again to find it's slope, then derived the second function to get it's slope, then i have set the first slope = the second slope anyway this is what i got lastly:

(3/4) x ^ (-1/2) + (1/4) x ^ (-3/2) = 1
i could not get the value of x from this equation.is there any formula that will get me the value of x like the quadratic formula?

The tangent line to y=x^(3/2)-x^(1/2) is parallel to y-x=3, when the
gradient is equal to that of the line.

The gradient of the line is 1, so you want the solutions of:

dy/dx=1.

RonL

thanks Captain for ur reply i have solved the question again and got the correct answear.anyway what if the derivative of the first function contained variable powers like this for example:
9/5 x^(4/5) - 8/3 x^(5/3) = 1
how can i find the value(s) of x i mean as there are two terms at the left side that contains different powers.(are there any formulas like the quadratic formula that will get me the value of x but with variable powers other than 2)??