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Camille's question at Yahoo! Answers regarding finding the parameter of a function

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MarkFL

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Feb 24, 2012
13,775
Here is the question:

Calculus tangent line?

If the line 3x-4y=0 is tangent in the first quadrant to the curve y=x^3+k then k is
(a)1/2
(b)1/4
(c)0
(d)-1/8
(e)-1/2
Could you please show work because I am so lost
I have posted a link there to this topic so the OP may see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Camille,

Let's begin by equating the derivative of the curve to the slope of the tangent line. The tangent line may be written as \(\displaystyle y=\frac{3}{4}x\) hence:

\(\displaystyle \frac{3}{4}=3x^2\)

\(\displaystyle x^2=\frac{1}{4}\)

Since we are interesting in the first quadrant solution, we take the positive root:

\(\displaystyle x=\frac{1}{2}\)

Now, the tangent line and the curve have then tangent point in common, and so using the tangle line, we know this point is:

\(\displaystyle \left(\frac{1}{2},\frac{3}{4}\cdot\frac{1}{2} \right)=\left(\frac{1}{2},\frac{3}{8} \right)\)

Hence, we must have:

\(\displaystyle y\left(\frac{1}{2} \right)=\left(\frac{1}{2} \right)^3+k=\frac{1}{8}+k=\frac{3}{8}\)

And so we must have:

\(\displaystyle k=\frac{1}{4}\)

Here is a plot of the curve \(\displaystyle y=x^3+\frac{1}{4}\) and the tangent line \(\displaystyle y=\frac{3}{4}x\) for $0\le x\le1$:

camille.jpg