Need Help with First/Second/Third Derivative relation

[tangents]

Hello all,

Well I am kind of stumped on this question and was hopeing for somr guidance.

X 1.1 1.2 1.3 1.4
f(x) 4.18 4.38 4.56 4.73

F is obviously a function and f ''(x)<o for all x in the closed interval. The question asks which of the following has to be for f '(1.2)?

1) f '(1.2)<0 This cant be it becuase f(x) is positive
2) 0<f '(1.2)<1.6
3) 1.6<f '(1.2) <1.8
4) 1.8< f '(1.2) <2.0
5) f '(1.2)> 2.0 This one is saying that the slope at 1.2 is greater than 2, but i suppose the only way to be certain is to graph the points and find out.


As for the rest I'm not entirely sure but hope you guys can assist me ; ;

 

[galactus]

Just for kicks, using your data I ran a Cubic regression and came up with:

\(\displaystyle y=\frac{5}{3}x^{3}-7x^{2}+\frac{689}{60}x-2.2\)

Check the equation and see when you enter in your x values you get the required y values.

\(\displaystyle \frac{dy}{dx}=5x^{2}-14x+\frac{689}{60}\)

\(\displaystyle \frac{dy}{dx}=5(\frac{6}{5})^{2}-14(\frac{6}{5})+\frac{689}{60}=\frac{113}{60}\)

I would opt for #4.

 

[tangents]

ah ok, What program did you use to do cubic regression? Also even though #4 looks close to the answer as shown by your data I however need to show work and I can't show the equation/derivative since the question never gave it to me

 

[galactus]

I used a TI-92 to find the cubic regression. It does all sorts of regressions.

You could find the slope on either side of 1.2 by using \(\displaystyle \frac{y-y_{1}}{x-x_{1}}\)

\(\displaystyle \frac{4.56-4.18}{1.3-1.1}=1.9\)

Pretty close to 1.88, what the derivative gave us.

If you're curious, here's something interesting.

A method for finding an equation which isn't too bad to follow.

Starting with your x data.

For x=1.1:

(x-1.2)(x-1.3)(x-1.4)=0 for x=1.2, 1.3, 1.4. It equals -.006 at x=1.1.

But we want it to equal 4.18 at x=1.1; 4.18/-.006=-2090/3

So, we have \(\displaystyle \frac{{-}2090}{3}(x-1.2)(x-1.3)(x-1.4)\)


Next, for x=1.2:

(x-1.1)(x-1.3)(x-1.4)=0 for 1.1, 1.3, 1,4. It equals .002 when x=1.2;

We want 4.38 at x=1.2; 4.38/.002=2190.

We have, \(\displaystyle 2190(x-1.1)(x-1.3)(x-1.4)\)


Next, x=1.3:

(x-1.1)(x-1.2)(x-1.4)=0 when x=1.1, 1.2, 1.4

It equals -.002 when x=1.3. We want 4.56 when x=1.3

4.56/-.002=-2280

We have: \(\displaystyle {-}2280(x-1.1)(x-1.2)(x-1.4)\)

Next, x=1.4:

(x-1.1)(x-1.2)(x-1.3)=0 when x-1.1, 1.2, 1.3

It equals .006 when x=1.4. We want 4.73 when x=1.4, 4.73/.006=2365/3

We have: \(\displaystyle \frac{2365}{3}(x-1.1)(x-1.2)(x-1.3)\)

Now, add them all up. You'll need a calculator.

\(\displaystyle \frac{-2090}{3}(x-1.2)(x-1.3)(x-1.4)+2190(x-1.1)(x-1.3)(x-1.4)-2280(x-1.1)(x-1.2)(x-1.4)+\frac{2365}{3}(x-1.1)(x-1.2)(x-1.3)\)

Simplified this equals:

\(\displaystyle \frac{5}{3}x^{3}-7x^{2}+\frac{689}{90}x-\frac{11}{5}\)

The same as the cubic regression gave us.