# Homework Help: Strange UBC approximation question

1. May 2, 2006

### Mr. Snookums

I'm in AP Calculuc and was given a homework package, which is an old university introductory calculus exam. There is one particular question with which I'm having a terrible time.

It is known that f(0)=5 and the tangent line to the graph of f(x) at (0,5) is y=5+3x. It is also known that abs(f''(x))<1/16 for all x. Determine a value of M (as small as possible) such that the error in using the tangent line as an approximation of f(x) on the interval [-2,2] is guaranteed to e smaller than M.

This is quite confusing. I don't even know what I should be approximating, and I don't know what M represents.

I know the rules say that we should post our attempts at the problem, but I honestly have no idea where to start other than drawing a diagram of this.

2. May 2, 2006

### 0rthodontist

You know abs(f''(x)) < 1/16, but what if it were equal to 1/16? What if it were equal to -1/16? What would be the maximum error in those cases?

3. May 2, 2006

### shmoe

M is an upper bound to |5+3x-f(x)| as x ranges over [-2,2]. |5+3x-f(x)| is the distance of this tangent line to the function at the point x, e.g. at x=0 we get 0 since f(0)=5.

You know f(0), so to get bounds on f(x) it would be enough to find f(x)-f(0). What would you need information about to bound a change in f like this?

4. May 4, 2006

### Mr. Snookums

Thank you for answering.

Is it assumed that when we see M in this type of question that it is the upper bound?

From where do we get abs(5+3x-f(x))?

My teacher told me it was a Taylor series problem, so the error would be:

abs((1/2)(yo'')(x-xo)^2)

Am I on the right track?

5. May 5, 2006

### HallsofIvy

Yes, that's right. The tangent line to y= f(x) at (x0,y0, y= f'(x0)(x- x0)+ y0, is just the first order Taylor's polynomial and so you can use the error formula for Taylor's polynomial. Don't use y0" for the second derivative however, for two reasons: first y0 is a number not a function so it doesn't make sense to differentiate it. I presume you really meant
y"(x0) or f"(x0). Second, even that's not correct. The second derivative should be evaluated at some unknown value between x and x0. Since the value is unknown, the best you can do is use a maximum value for the second derivative in that interval.
Fortunately, you are told "abs(f''(x))<1/16 for all x."

6. May 5, 2006

### shmoe

That would be the absolute error between the function and the line you are using to approximate it. By the way, my post was assuming you didn't know Taylor polynomials (with relevant error theorems) and had to derive an error "from scratch". Since you do know this what you've done is fine, given what Halls has pointed out.

7. May 5, 2006

### Mr. Snookums

So now I just plug 1/16 in for f''(xo) and make M greater than it? What about x?

M>x^2/32.

Is x any number in the interval [-2,2]?