arianna1453
  • arianna1453
FAN +MEDAL Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is g′(x) = x^2-16/x-2 with x ≠ 2. Find all values of x where the graph of g has a critical value. For each critical value, state whether the graph of g has a local maximum, local minimum, or neither. You must justify your answers with a complete sentence. On what intervals is the graph of g concave down? Justify your answer. Write an equation for the tangent line to the graph of g at the point where x = 3.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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arianna1453
  • arianna1453
1. I got the critical values of x= 4 , -4 2. I got that since g''(x) >0 for x=4 , -4 that they are both local minimums. 3. I am stuck on finding the intervals of concave down.
arianna1453
  • arianna1453
@Astrophysics Can you help?
arianna1453
  • arianna1453
@mathmale ?

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arianna1453
  • arianna1453
@ParthKohli
mathmale
  • mathmale
"x ≠ 2" ?? Why the focus on x=2, if the function is defined for all x?
mathmale
  • mathmale
This statment, "g′(x) = x^2-16/x-2 with x ≠ 2," is unusual. Why "with x=2?"
arianna1453
  • arianna1453
x cannot equal 2. Theres a vertical asymtote there.
mathmale
  • mathmale
the important thing here is that you're given the derivative of function g(x). How do you go about using this derivative to determine the critical values of g(x)?
mathmale
  • mathmale
OK. then write \[x \neq2\]
arianna1453
  • arianna1453
Ive already figure out the critical values. Its 4, -4
mathmale
  • mathmale
This is the opposite of x=2, which was the reason I needed to ask for clarification.
mathmale
  • mathmale
Please demonstrate that 4 and -4 are indeed critical values of g(x). How would you do that?
arianna1453
  • arianna1453
You set the derivative equal to 0. then solve for x. leading you to obviously get 4, -4
mathmale
  • mathmale
So, if you substitute x=4 into g'(x), you should obtain what result?
arianna1453
  • arianna1453
You would get 0.
mathmale
  • mathmale
Great. Let's move on.
mathmale
  • mathmale
What do you need help with?
arianna1453
  • arianna1453
On what intervals is the graph of g concave down? Justify your answer.
mathmale
  • mathmale
Starting with the given g '(x), find the second derivative, g ''(x).
arianna1453
  • arianna1453
\[g''(x)= \frac{ x ^{2}-4x+16 }{ (x-2)^{2} }\]
mathmale
  • mathmale
I'm not going to check that. Given that you started with a rational function, your obtaining a rational function as the second derivative is approrpriate. Please set this second derivative = to 0 and solve for x. This is to set up intervals on which g(x) is concave up and on which it's concave down.
mathmale
  • mathmale
Hints: If you set this g ''(x)=0 and want to solve for x, that's equivalent to setting the numerator = 0 and solving for x. Note that g ''(x) is not definied at x=2. Plot x=2 on a number line. If the roots of g ''(x) are real, also graph them on the same number line.
arianna1453
  • arianna1453
Thats the part I dont understand.
arianna1453
  • arianna1453
2+2i sqrt3 ? like i dont understand
mathmale
  • mathmale
Set the numerator of this rational function = to 0. In other words,\[x^2-4x+16=0\] I see you've already done this and have solved for the roots. Your roots are "complex". No point in graphing them on the real x axis.
mathmale
  • mathmale
the only relevant real number to graph on the x-axis is x=2. Choose an x value smaller than 2 and then another one greater than 2. Substitute these 2 numbers into the 2nd derivative, only for the purpose of determining the sign of the 2nd derivative at each value of x. Results?
arianna1453
  • arianna1453
I chose 3 and 1. Both god positive 13
mathmale
  • mathmale
On the interval\[(-\infty,2),\] the second derivative is positive? negative? On the interval \[(2m \infty), the second der. is positive? \neg?\]
mathmale
  • mathmale
So, your 2nd derivative is + on both sides of x=2. What does that tell you about the direction of concavity of each half of the graph?
arianna1453
  • arianna1453
Its concave up.
arianna1453
  • arianna1453
which means since the second derivative is always positive, it is never concave down, correct?
mathmale
  • mathmale
yes. very good. However, I'd prefer you mention that x=2. the sec. deriv. is undefinited there, so I wouldn't want to say the curve of g(x) is concave up at x=2. Instead, use sets, as I did above.
arianna1453
  • arianna1453
Okay! Awesome. For the next step, Write an equation for the tangent line to the graph of g at the point where x = 3. Would it be: y-y1=m(x-x1) y-g(3) = g'(3)(x-3) y-4=-7(x-3)
mathmale
  • mathmale
Yes, very good. You already have a formula for the slope of the t. l, and need only subst. x=3 into it to determine the slope of the tan. line at x=3. How do you propose to find the value of g(3)?
arianna1453
  • arianna1453
g(3) = 4 with the given information that the problem gave us. Then to find g'(3) I substituted 3 for x in g'(x) and got -7.
mathmale
  • mathmale
I haven't attempted that. But you demonstrate considerable competence, so I'll go along with your result here.
arianna1453
  • arianna1453
so plugging it all into the equation, so far i got y-4 = -7 (x-3)
mathmale
  • mathmale
Again, that strikes me as appropriate , although I haven't done the actual calculations myself. Nice going on your part!
arianna1453
  • arianna1453
Would I need to get it into y=mx+ form? or would that be the equation for the tangent line at x=3?
arianna1453
  • arianna1453
hello?
mathmale
  • mathmale
that's be an appropriate form, the simplest you could choose. Sorry for the delay; I got sidetracked. Are you comfortable with the solution you've reached?
arianna1453
  • arianna1453
Yes, completely perfect, Thank you!

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