MHB Find the coordinates and the nature of any turning points (maximum and minimum)

[b4rn5ey]

So I've been trying to wrap my head around this one for several hours now and it just has me stumped. I'm begging, someone, anyone, walk me through it before I swear off numbers for life hahah (Part b)

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Thanks in advance for any help anyone can provide :)

 

[MarkFL]

Hello, and welcome to MHB, b4rn5ey! (Wave)

In part (a) of the problem, you correctly found:

$$y'=16x^3-18x^2+4x$$

To find the turning points, and we observe that $y'=0$ at such points, we need to find the roots of $y'$ (our critical values)...that is, we need to solve:

$$16x^3-18x^2+4x=0$$

I would suggest factoring to get a linear factor and a quadratic factor...what do you get?

 

[b4rn5ey]

Hello, and welcome to MHB, b4rn5ey! (Wave)

In part (a) of the problem, you correctly found:

$$y'=16x^3-18x^2+4x$$

To find the turning points, and we observe that $y'=0$ at such points, we need to find the roots of $y'$ (our critical values)...that is, we need to solve:

$$16x^3-18x^2+4x=0$$

I would suggest factoring to get a linear factor and a quadratic factor...what do you get?

Thanks for the reply..

So I've tried to find the first turning point using your advice..

This seem right? Or am i walking off track>

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View attachment 6601

 

[MarkFL]

I agree with your factorization:

$$y'=2x\left(8x^2-9x+2\right)=0$$

I also agree with the critical values you obtained from the quadratic factor:

$$x=\frac{9\pm\sqrt{17}}{16}$$

However, there is another factor involving $x$ that must be equated to zero, which is simply:

$$x=0$$

So, this gives us 3 critical values, and 3 possible turning points:

$$y(0)=-1$$

$$y\left(\frac{9-\sqrt{17}}{16}\right)=\frac{51\sqrt{17}-2155}{2048}$$

$$y\left(\frac{9+\sqrt{17}}{16}\right)=-\frac{51\sqrt{17}+2155}{2048}$$

Now, you need to determine the nature of these points, using either the first or second derivative tests. :D

If you are going to upload images of your work rather than typeset it for easier reading, please attach the images inline rather than hot-linking to the image...this way I won't have to follow the link, download the image, resize and crop the image, and edit your post to attach it inline so it can be read in the post.