sketch the graph of y =x^4 +3 (done this)

2. show that y = 4x is a tangent to this curve and find the point of contact

3. sketch this tangent on your original set of axis

please someone tell me what to do

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- Oct 2nd 2009, 08:24 PM #1

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- Oct 2nd 2009, 08:29 PM #2
Taking the derivative of the function gives us $\displaystyle y'=4x^3$. Now the line $\displaystyle y=4x$ has slope $\displaystyle 4$, so we want to find which $\displaystyle x$ will give us $\displaystyle y'=4$. Solving for $\displaystyle x$:

$\displaystyle 4x^3=4 \implies x=1$

So the point of contact is $\displaystyle x=1$ (and $\displaystyle y=4$).