Solving a Math Problem: Find k and y-coordinate of Point A on Curve C

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Hi everyone,
I have stumbled upon this maths problem, i have pondered it for a few days without any real work until today. The problem is, i can't seem to comprehend these questions at all... so any help would very much appreciated :biggrin:

Here is the math problem:

The curve C has equation y=k*x^3-x^2+x-5, where k is constant.

A) Find the derivative of the expression with respect to x

I got 3*k*x^2-2 x+1=y

The point A with x-coordinate -1/2 lies on c. The tangent c at a is parallel to the line with equation 2*y-7*x+1=0

Find:

B) the value of k

C) the value of the y-coordinate of A.
 
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jaycool1995 said:
Hi everyone,
I have stumbled upon this maths problem, i have pondered it for a few days without any real work until today. The problem is, i can't seem to comprehend these questions at all... so any help would very much appreciated :biggrin:

Here is the math problem:

The curve C has equation y=k*x^3-x^2+x-5, where k is constant.

A) Find the derivative of the expression with respect to x

I got 3*k*x^2-2 x+1=y
The derivative is correct, but you have mislabelled it as "y". You should have y' = 3kx^2 - 2x + 1 or dy/dx = 3kx^2 - 2x + 1.
jaycool1995 said:
The point A with x-coordinate -1/2 lies on c. The tangent c at a is parallel to the line with equation 2*y-7*x+1=0
Point A lies on C. The tangent to C at A is parallel to a certain line.

The derivative you found can be used to find the slope of the tangent line to C at any point. What's the slope of the tangent line (on C) at the point A? What's the slope of the line whose equation is given?
jaycool1995 said:
Find:

B) the value of k

C) the value of the y-coordinate of A.
 
So, as far as i understand it, the slope of the line that is tangent to C at point A (-1/2). The second equation given is parallel to the first derivative of C when x=-1/2
My first thought was to do this:
3 k x^2-2 x+1=2 y+\frac{9}{2}
which is the same as

y'\left(-\frac{1}{2}\right) =2 y+\frac{9}{2}
which is equal to the line that is tangent to C. From this you can solve for k, which i got k=2. But i am struggling on how to find y, how could you use the other equation to find y, when the line is the same but rather only parallel. How could i find y in this case?
So wouldn't the equation look like this with the different y's:
\left(y_1\right)'\left(-\frac{1}{2}\right)\to \frac{3 k}{4}+2=2 y_2+\frac{9}{2}
or can we find y by going back to the original equation like this:
y=2 \left(-\frac{1}{2}\right)^3-\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)-5=-6
This is why it is confusing me lol i'll get there eventually ;)
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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