Applications of the first derivative. Extrema?

[cmnalo]

I need to find the relative extrema,if any, of the function.

f(x) = x^3 -X^2 -5x +6
f'(x) = 3x^2 - 2x -5
f'(x) = (3x-5)(1x+1)
3x-5=0 1x+1=0
x=5/3 x= -1

(-infinity, -1) (-1, 5/3) (5/3, infinity)
test point x= -2 x=1 x=2
Sign of f'(x) f'(-2)= 11>0 f'(x) =-4 <0 f'(x) = 3>0
conclusion Increasing Decreasing Incresing

The answer is:
Rmin: f(5/3) = 13/27
Rmax: f(-1)= 9

I don't understand how the answer is arrived at and where might be going wrong. Any help would be greatly appreciated.

 

[soroban]

Re: Applications of the first derivative

Hello, cmnalo!

What you did was correct . . . you just didn't recognize it.

Find the relative extrema of the function: \(\displaystyle \,f(x) \:= \:x^3\,-\,x^2\, -\,5x\, \,+6\)

\(\displaystyle f'(x) \:=\: 3x^2\,-\,2x\,-\,5\)

\(\displaystyle f'(x) \:= \3x\,-\,5)(x\,+\,1)\:=\:0\)

\(\displaystyle \begin{array}{cc}3x\,-\,5\:=\:0 & \;\; & \;\;x\,+\,1\:=\:0 \\x\,=\,\frac{5}{3} & \;\; & x\,=\,-1\end{array}\)


\(\displaystyle \begin{array}{cccc}\text{Intervals: }\\ \text{Test point: }\\ \text{Sign of f'(x): } \\ \text{Conclusion: }\end{array}\;\;\;
\begin{array}{cccc}(-\infty,\,-1) \\x\,=\,-1 \\ f'(-1)\,=\,+11 \\ \nearrow \end{array}\;\;\;
\begin{array}{ccc}\left(-1,\,\frac{5}{3}\right) \\ x\,=\,1 \\ f'(1)\,=\,-4 \\ \searrow \end{array}\;\;\;
\begin{array}{ccc}\left(\frac{5}{3},\,\infty\right) \\ x\,=\,2 \\ f'(2)\,=\,+3 \\ \nearrow\end{array}\)
. . . . . . . . . . . . . . . . . . . . \(\displaystyle \fbox{x\,=\,-1}\;\;\;\;\;\fbox{x\,=\,\frac{5}{3}}\)

Do you see what you've discovered?

To the left of \(\displaystyle x\,=\,-1\), the graph is increasing.
At \(\displaystyle x \,=\,-1\), there is a horizontal tangent.
To the right of \(\displaystyle x\,=\,-1\), the graph is decreasing.
. . The graph is shaped like this: \(\displaystyle \;\nearrow\;^{\longrightarrow}\,\searrow\)
. . Hence, there is a Rmax at: \(\displaystyle x\,=\,-1,\;y\,=\,9\)

To the left of \(\displaystyle x\,=\,\frac{5}{3}\), the graph is decreasing.
At \(\displaystyle x\,=\,\frac{5}{3}\), there is a horizontal tangent.
To the right of \(\displaystyle x\,=\,\frac{5}{3}\), the graph is increasing.
. . The graph is shaped like this: \(\displaystyle \;\searrow\;_{\longrightarrow}\,\nearrow\)
. . Hence, there is a Rmin at: \(\displaystyle x\,=\,\frac{5}{3},\;y\,=\,-\frac{13}{27}\)

Got it?

 

[cmnalo]

I really appreciate the help. You make things very clear. However, I'm not totally clear on how the y=9 and y=-13/27 is arrived at. I thought you plug the -1 and 5/3 back into the derivative??

 

[stapel]

The max/min points are point on the original curve. Therefore, to find the y-values for these points, one must plug the x-values into the original curve.

Eliz.