....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)?
....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 $\displaystyle 3x^2-12$Quote:
Originally Posted by plstevens
Setting it equal to zero, we get $\displaystyle x^2-4=0\implies x^2=4\implies x={\color{red}\pm}2$
These are the only values where the function has horizontal tangents.
Does this make sense?
--Chris
Hi to find a horizontal tangent line to a function you take the derivative and equal it to 0
$\displaystyle f(x) = x^3 -12x+2$
$\displaystyle f'(x) = 3x^2 -12$
$\displaystyle 0= 3x^2 -12$
$\displaystyle x = 2$ and $\displaystyle x = -2$
Shoot to late again lol
thanks
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
the computer keep saying its wrong, where did i possibly go wrong on this one
$\displaystyle 9x^{\frac{7}{5}} - 5x^2+10^4$
Is this what it looks like?
