Find all the values of x (if any) where the tangent line to the graph of the....

**plstevens** · September 3rd 2008, 09:53 PM

....function is horizontal.
x cubed - 12x + 2

wouldn't that be 0 and 2. by taking the derivative and solving for x, i get only 2, how do i find the other number(s)?

**Chris L T521** · September 3rd 2008, 09:56 PM

Originally Posted by plstevens

....function is horizontal.
x cubed - 12x + 2

wouldn't that be 0 and 2. by taking the derivative and solving for x, i get only 2, how do i find the other number(s)?

The derivative is $3x^2-12$

Setting it equal to zero, we get $x^2-4=0\implies x^2=4\implies x={\color{red}\pm}2$

These are the only values where the function has horizontal tangents.

Does this make sense?

--Chris

**11rdc11** · September 3rd 2008, 10:02 PM

Hi to find a horizontal tangent line to a function you take the derivative and equal it to 0

$f(x) = x^3 -12x+2$

$f'(x) = 3x^2 -12$

$0= 3x^2 -12$

$x = 2$ and $x = -2$

Shoot to late again lol

**plstevens** · September 3rd 2008, 10:03 PM

thanks

**plstevens** · September 3rd 2008, 10:14 PM

9x^7/5-5x^2+10^4

Now here i worked this out but the computer keep saying my answer of 63/5x^2/5-10x

**plstevens** · September 3rd 2008, 10:14 PM

the computer keep saying its wrong, where did i possibly go wrong on this one

**11rdc11** · September 3rd 2008, 10:46 PM

$9x^{\frac{7}{5}} - 5x^2+10^4$

Is this what it looks like?

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