Asked by Justin
We are finding critical numbers
f(x)= 8x^3+x^2+8x
f'(x)=24x^2+2x+8 -> 2(12x^2+x+4)
I'm honestly stuck. All I can tell is that I need to set my f'(x)=0, but I am lost there too.
Quadratic form. equates to (-1±i√191)/24 I believe. That is as far as I can get.
f(x)= 8x^3+x^2+8x
f'(x)=24x^2+2x+8 -> 2(12x^2+x+4)
I'm honestly stuck. All I can tell is that I need to set my f'(x)=0, but I am lost there too.
Quadratic form. equates to (-1±i√191)/24 I believe. That is as far as I can get.
Answers
Answered by
Steve
so, what's the trouble? You have critical numbers where f'=0. But, as you found out, f' is never zero.
So, ... there are no critical numbers. There is no place where the tangent line is horizontal. Just look at the graph and sigh:
http://www.wolframalpha.com/input/?i=8x^3%2Bx^2%2B8x
So, ... there are no critical numbers. There is no place where the tangent line is horizontal. Just look at the graph and sigh:
http://www.wolframalpha.com/input/?i=8x^3%2Bx^2%2B8x
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