1. ## last function problem

Show that the graph f(x)=(x^3)+(2x^2)+6x does not have a tangent line with a slope of 4.

Im not really sure where to start with this one...find f'(x) first or what?

2. Originally Posted by whoislambo
Show that the graph f(x)=(x^3)+(2x^2)+6x does not have a tangent line with a slope of 4.

Im not really sure where to start with this one...find f'(x) first or what?
Hi

Yes , you differentiate first to find the gradient of the tangent line .

$\displaystyle f'(x)=3x^2+4x+6$

Now check its discriminant

$\displaystyle b^2-4ac=4^2-4(3)(6)=-56<0$

Since its smaller than 0 , hence it has no real roots , which means no value of x would give you 4 , which is the gradient of the tangent line .

3. Originally Posted by whoislambo
Show that the graph f(x)=(x^3)+(2x^2)+6x does not have a tangent line with a slope of 4.

Im not really sure where to start with this one...find f'(x) first or what?
Find $\displaystyle f'(x)$ and show it can never equal $\displaystyle 4$.

$\displaystyle f'(x)=3x^2+4x+6$

Show $\displaystyle 3x^2+4x+6=4$ has no real solutions.

Spoiler:
Use the quadratic formula on $\displaystyle 3x^2+4x+2=0$, so $\displaystyle a=3$, $\displaystyle b=4$, and $\displaystyle c=2$.

$\displaystyle \frac{-4\pm\sqrt{16-24}}{6}=\frac{-4\pm\sqrt{-8}}{6}=-\frac{2}{3}\pm \frac{\sqrt{2}}{3}i$

So there are no real roots, and therefore the slope of the tangent line never equals $\displaystyle 4$. (You actually could have stopped as soon as you saw that the discriminant in the quadratic formula was negative, but I figured I'd finish it out to the end.)