Finding relative extrema of a cubic function

[Nikolas7]

Find local maxima and minima for 6${x}^{3}$+6${x}^{2}$-8x. I found that (-1.08,8.08) is max, (0.41,-1.86) is min. Where i was wrong?

 

[MarkFL]

This problem is an application of differential calculus, and so I have moved your thread to the appropriate forum. For better forum organization, we ask this in MHB RUle #5:

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Find local maxima and minima for 6${x}^{3}$+6${x}^{2}$-8x. I found that (-1.08,8.08) is max, (0.41,-1.86) is min. Where i was wrong?

Without seeing your work, we cannot possibly tell you where you went wrong. What did you get when you differentiated the given function in preparation for equating it to zero?

 

[Nikolas7]

18${x}^{2}$+12x-8=0 It's quadratic equation. Solve, I got x1=0.41 x2=-1.08. Then I found max and min 8.08 and -1.86.
Thank you.

 

[MarkFL]

We are given:

$$f(x)=6x^3+6x^2-8x=2\left(3x^3+3x^2-4x\right)$$

Hence:

$$f'(x)=2\left(9x^2+6x-4\right)=0$$

Applying the quadratic formula, we find:

$$x=\frac{-6\pm\sqrt{6^2-4(9)(-4)}}{2(9)}=\frac{-1\pm\sqrt{5}}{3}$$

Your roots are correct to 2 decimal places. Now, when we look at the first derivative, we see it is a quadratic, a parabola opening upwards with 2 read roots. This means it is positive to the left of the smaller root, negative in between the roots, and positive to the right of the larger root. What does this then tell us about the nature of the extrema at our two critical values?

 

[Nikolas7]

at smaller root we have local maximum and at bigger root we have local minimum. Is it right?

 

[MarkFL]

at smaller root we have local maximum and at bigger root we have local minimum. Is it right?

Yes, we know the slope of $f$ is positive to the left of the smaller turning point and negative to the right of if, then negative to the left of the larger turning point and positive to the right, so your conclusion is correct. So you just need to evaluate:

$$f_{\max}=f\left(\frac{-1-\sqrt{5}}{3}\right)$$

and

$$f_{\min}=f\left(\frac{-1+\sqrt{5}}{3}\right)$$

 

[Nikolas7]

Thanks, I did too and got points: max (-1.08,8.08), min (0.41,-1.86).

 

[MarkFL]

Thanks, I did too and got points: max (-1.08,8.08), min (0.41,-1.86).

Those are correct to two decimal places. The exact values are:

$$f_{\max}=\frac{4}{9}\left(7+5\sqrt{5}\right)$$

$$f_{\min}=\frac{4}{9}\left(7-5\sqrt{5}\right)$$

 

[Nikolas7]

Thanks

 

[offirgo]

Find local maxima and minima for 6${x}^{3}$+6${x}^{2}$-8x. I found that (-1.08,8.08) is max, (0.41,-1.86) is min. Where i was wrong?

check this out, will help