Question

What is the relative maximum and minimum of the function?
f(x)= 2x^3 + x^2 – 11x


The relative maximum is at (–1.53, 8.3) and the relative minimum is at (1.2, –12.01).

The relative maximum is at (–1.53, 12.01) and the relative minimum is at (1.2, –8.3).

The relative maximum is at (–1.2, 8.3) and the relative minimum is at (1.53, –12.01).

The relative maximum is at (–1.2, 12.01) and the relative minimum is at (1.53, –8

Answer 1

first of all, will you be using an algebra approach or a calculus approach to answer this question?

Answer 2

algebra

Answer 3

In that case your best bet is to graph the function. It'll take a while. But once you've graphed it, you'll be able to spot both the min and the max easily.

Answer 4

calculus would be much quicker, but you and I may have to save that for later.

Answer 5

ohhhhh wait

Answer 6

Its D, right? I did graph it but i didnt know if there was an equation i had to do

Answer 7

Need to see your graph before I'll offer an opinion on your choice of D.
Could you possibly do a screen shot and share it?

f(x)= 2x^3 + x^2 – 11x

Answer 8

Yes, hold on! :)

Answer 9

Note that f(x)= 2x^3 + x^2 – 11x could be at least partially factored.

Answer 10

@mathmale

Answer 11

@mathmale

Answer 12

@Michele_Laino

Answer 13

we have to compute the first derivative of such function

Answer 14

how/

Answer 15

it is simple, we can apply this formula:
\[\left( {{x^n}} \right)' = n{x^{n - 1}}\]
where at the left side I have indicated the first derivative of \(x^n\)
then we can compute the first derivative of the function, term by term.
So, we get:
\[\left( {2{x^3} + {x^2} - 11x} \right)' = 6{x^2} + 2x\]

Answer 16

oops.. i have made an error:
\[\Large \left( {2{x^3} + {x^2} - 11x} \right)' = 6{x^2} + 2x - 11\]

Answer 17

now, please solve this inequality:
\[6{x^2} + 2x - 11 > 0\]

Answer 18

27 > 0?

Answer 19

we have to solve this equation first:
\[6{x^2} + 2x - 11 = 0\]

Answer 20

thats what i did

Answer 21

27>0 is the answer

Answer 22

hint:
the solution of such equation are:

\[\Large {x_1} = \frac{{ - 1 + \sqrt {67} }}{2},\quad {x_2} = \frac{{ - 1 - \sqrt {67} }}{2}\]
so, the solution of the above inequality is:
\(x<x_1\) and \(x>x_2\)

Answer 23

please note that where \(f′(x)>0\) then function \(f(x)\) is an increasing function

Answer 24

so, we have this drawing:

Answer 25

If you're attempting to solve this problem graphically, then the graph yuo've shared is perfect for the job. You can see both the min and the max of the given function on this graph. Find the approx coordinates of the min and those of the max.

Dani: You say your approach is "algebraic." In this case, graphing is the best approach to finding the min. and the max. of the given function.

Michele_Laino's approach uses calculus, which is a powerful tool, but not if you haven't already studied some calculus.

any questions?

Answer 26

sorry, the solution of the above inequality is:
\(x<x_2\) union \(x>x_1\)

Answer 27

Dani: Once again: Have you studied any calculus? If so, are you familiar with the concept of "derivative?"

Answer 28

so I can update the drawing above:

and now, we know what is the point of maximum, and what is the point of minimum