The function f(x)=2x^{3}-33x^{2}+168x+9 has one local minimum and one local

The function $f\left(x\right)=2{x}^{3}-33{x}^{2}+168x+9$ has one local minimum and one local maximum.
Use a graph of the function to estimate these local extrema.
This function has a local minimum at $x=?$
with output value:
and a local maximum at $x=?$
with output value:
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Muspee
Step 1
To estimate (approximate calculation) the local maximum and local minimum of the given function
$y=f\left(x\right)$
Step 2
Now we know the critical values are at $x=4$ and $x=7$. For further analysis graph the function between $x=3$ and $x=8$ (in the neighbourhood of the critical points)
$f\left(x\right)=2{x}^{3}-33{x}^{2}+168x+9$
The local max and min occur when
${f}^{\prime }\left(x\right)=6{x}^{2}-66x+168$
$=6\left({x}^{2}-11x+28\right)$
$=6\left(x-4\right)\left(x-7\right)=0$
so,
Step 3
Estimates: local minimum at $x=7$ with value 255, local maximum at $x=4$ with value 280