Respuesta :

Find the first derivative, using the power rule, as follows

[tex]f'(x) = 3x^2 + 2x - 1[/tex]

We know that a max/min of a polynomial function will always occur when [tex]f'(x) = 0[/tex].  

Therefore, we must solve the equation 

[tex]3x^2 + 2x - 1 = 0[/tex]

We can factor this 

[tex]3x^2 + 3x - x - 1 = 0 \\ 3x(x + 1) - (x +1) = 0 \\ (3x - 1)(x + 1) = 0 \\ x = 1/3, -1 [/tex]

Now we see that at [tex]x = 0[/tex] the derivative is negative, at [tex]x =-2[/tex] the derivative is positive and at [tex]x = 1[/tex] the derivative is also positive.   Therefore, [tex]x = -1[/tex] is a relative maximum and [tex]x =1/3[/tex] is a relative minimum.  

I will leave it as an exercise to determine the corresponding values of y. However we can confirm graphically that our points are the maximum and minimum.  



Hopefully this helps!
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Answer:

At x=-1 is a local maximum.

At x=0.33 is a local minimum.

Step-by-step explanation:

Given : Function [tex]f(x)=x^3+x^2-x-1[/tex]

To find : Use the graph of the function to identify its relative maximum and minimum.            

Solution :

First we plot the graph of the function [tex]f(x)=x^3+x^2-x-1[/tex]

Refer the attached figure below.

We know, When the curve is concave down then it gives you maximum and if the curve is concave up then it gives you minimum.

When we examine the graph we get,

The curve is concave down at point (-1,0) and

The curve is concave up at point (0.33,-1.18).

Therefore,

At x=-1 is a local maximum and

At x=0.33 is a local minimum.

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