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Find the slope of f(x) = x 3−3x at the point x = 2 and the corresponding equation of the tangent line. Use the definition of the slope of graph at a point. 

I got to lim h=0 (x+h)^3-3(x+h)-(x^3-3x)/h)

 Apr 1, 2021

Best Answer 

 #1
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$\lim_{h\to 0}\frac{(x+h)^3-3(x+h)-x^3+3x}{h}=\lim_{h\to 0}\frac{-3 h + h^3 + 3 h^2 x + 3 h x^2}{h}=\lim_{h\to 0} -3+h^2+3hx+3x^2=3x^2-3$

At $x=2$, this is $3*4-3=9$.

 

Using this, you can easily find the tangent line equation so I will leave that to you.

 Apr 1, 2021
 #1
avatar+594 
+2
Best Answer

$\lim_{h\to 0}\frac{(x+h)^3-3(x+h)-x^3+3x}{h}=\lim_{h\to 0}\frac{-3 h + h^3 + 3 h^2 x + 3 h x^2}{h}=\lim_{h\to 0} -3+h^2+3hx+3x^2=3x^2-3$

At $x=2$, this is $3*4-3=9$.

 

Using this, you can easily find the tangent line equation so I will leave that to you.

thedudemanguyperson Apr 1, 2021

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