Question

Check my asnwer please

What is the relative maximum and minimum of the function?
f(x)= x^3 + 2x^2 – 18x

A. The relative maximum is at (3, 22), and the relative minimum is at (–2, –41).
B. The relative maximum is at (–3, 41), and the relative minimum is at (2, –22).
C. The relative maximum is at (3, 41), and the relative minimum is at (–2, –22).
D. The relative maximum is at (–3, 22), and the relative minimum is at (2, –41).

I think it's B. Am I correct or not ?

Answer 1

You have:
\[f(x)=x^3+2x^2-18x\]
So take the derivative of it:
\[f'(x)=3x^2+4x-18\]
So then we solve for the roots of this quadratic and then test it with the second derivative rule to find out if it is a max or min

Answer 2

\[f \prime \left( x \right)=3x ^{2}+4x-18\]
\[3x ^{2}+4x-18=0,x=\frac{ -4\pm \sqrt{4^{2}-4*3*-18} }{2*3 }=\frac{ -4\pm \sqrt{16+216} }{ 6 }\]
\[x=\frac{ -4\pm2\sqrt{58} }{ 6 }=\frac{ -2 }{3 }\pm \frac{ \sqrt{58} }{3 }\]
now check for these two points.
I think there is problem with your statement.
No interval is given ,we have to check only these two points.

Answer 3

So it's not B ?

Answer 4

points are not given in your options.

Answer 5

check your question.

Answer 6

That's how the question exactly is, I checked.

Answer 7

\[f \prime \prime \left( x \right)=6x+4\]
\[put x=\frac{ -2 }{ 3 }+\frac{ \sqrt{58} }{3 } in f \prime \prime \left( x \right)\]

\[x=\frac{ -2+7.61 }{3 }=\frac{ 5.61 }{3 }\]
\[f \prime \prime \left( x \right)=6*\frac{ 5.61 }{3 }+4>0\]
so there is relative minima at this point.
\[when x=\frac{ -2-7.61 }{3 }=\frac{ -9.61 }{3 }\]
at this point
\[f \prime \prime \left( x \right)=6*\frac{ -9.61 }{3 }+4<0 \]
Hence there is relative maxima at this point.