# Can you check my solution? (limits)

• February 11th 2008, 05:59 PM
theowne
Can you check my solution? (limits)
3) Find the equation of the tangent line to y=x[squared] - 2 at the point where x=2

My solution:

slope: (change in y over change in x)

f(x) - f(2)
----------
x-2

which after substitution equals

x[squared]-4
------------
x-2

Factor and cancel out to leave [x+2]

Then find slope at [2]

lim[x->2] = 2+2 = 4

Now, find equation:

y-y0 = m(x-x0)

f(x)-f(2) = m (x-2)
x[squared] - 4 = 4(x-2)
x[squared]-4x+4

Equation is x[squared]-4x+4
• February 11th 2008, 07:50 PM
angel.white
Quote:

Originally Posted by theowne
3) Find the equation of the tangent line to y=x[squared] - 2 at the point where x=2

My solution:

slope: (change in y over change in x)

f(x) - f(2)
----------
x-2

which after substitution equals

x[squared]-4
------------
x-2

Factor and cancel out to leave [x+2]

Then find slope at [2]

lim[x->2] = 2+2 = 4

Now, find equation:

y-y0 = m(x-x0)

f(x)-f(2) = m (x-2)
x[squared] - 4 = 4(x-2)
x[squared]-4x+4

Equation is x[squared]-4x+4

Quick summary of how this will work. We will find the derivative of the function, then use this to find the slope of the point where x=2. Then we know that this line goes through the point (2,f(2)) so we will find that point, and plug those values and the slope into the equation of a line to find the y-intercept, then we will have everything we need for the equation of the tangent line. It's slope will be the slope of the derivative at the point x=2, and by selecting the appropriate value for b, we will guarantee that it goes through the correct points.

$f(x)=x^2-2$

Find the point that the tangent line will pass through:
$f(2)=2^2-2=4-2=2$

So it passes through the point (2,2)

I use a different way to find the derivative than you, I like this way better, it just makes more sense to me. Basically it says $\frac{f(x)-f(x)}{x-x}$ is the ideal way to find the slope of the function itself, but this of course leads to 0/0. So choose a value that is approaching zero (typically "h" is chosen to represent this) and add it to the function. So make one of the x into x+h, it really doesn't matter which you choose, heck, you could even use x-h if you wanted, it is merely showing that there is a very slight difference between the two f(x) values. So then the equation $\frac{f(x)-f(x)}{x-x}$ becomes $\frac{f(x)-f(x+h)}{x-(x+h)}$ or $\frac{f(x+h)-f(x)}{(x+h)-x}$. I used the first one, but I think the second one would have been simpler. Then what you are basically doing is finding the change in y over the change in x for the smallest possible increment.

$f\prime(x) = \lim_{h\to 0} \frac{f(x) - f(x+h)}{x-(x+h)}$

$f\prime(x) = \lim_{h\to 0} \frac{f(x) - f(x+h)}{-h}$

$f\prime(x) = \lim_{h\to 0} \frac{x^2-2 - (x+h)^2+2}{-h}$

$f\prime(x) = \lim_{h\to 0} \frac{x^2 - (x+h)^2}{-h}$

$f\prime(x) = \lim_{h\to 0} \frac{x^2 - x^2 -2xh-h^2}{-h}$

$f\prime(x) = \lim_{h\to 0} \frac{-2xh-h^2}{-h}$

$f\prime(x) = \lim_{h\to 0} 2x+h$

$f\prime(x) = 2x+0 = 2x$

So the slope of the function line is 2x. We can find the slope at any point now that we have this formula. So we want the slope at the point where x=2, so we can find this by plugging 2 into the equation:
$f\prime(2) = 2(2) = 4$

So 4 is the slope of the tangent line at x=2. This is m in the equation y=mx+b and we know the y and the x because we know it must pass through the point (2,2)

$y=mx+b$
$2=4*2+b$
$2=8+b$
$-6=b$

So now just fill in the values:
y=mx+b
y=4x-6

This is the slope of the tangent line.

edit: If all of your equations are relatively simple like this, I can tell you an easy way to cheat. It is called the power rule, it says that the derivative of $f(x)=cx^n$ where c is any coefficient, and n is any power, then the derivative will be $f\prime(x)=cnx^{n-1}$ So if you want the derivative of our function $f(x)=x^2-2$ the answer will be $f\prime(x)=2x^{2-1}=2x$ which you will notice is what we came up with. For a function such as $f(x)=5x^4+5x^3+5x^2+5x+5$ That since they are being added (note that it does not work this way with multiplication) they can each be differentiated separately so it will become $f\prime(x)=5(4)x^{4-1}+5(3)x^{3-1}+5(2)x^{2-1}+5(1)x^{1-1}+5(0)x^{0-1}$ which will simplify to $f\prime(x)=20x^3+15x^2+10x+5$. So at the very least, you can use that to check your work, and make sure you got the proper slope. (remember this gives the slope of the function, if you want to find the slope at the point you must plug in the value of x). And then you can check to make sure your equation for the tangent line is correct by plugging in your x value, your slope, and your y-intercept, and if they are all correct, it should return the value of y that you get when you plug x into the equation of your function. For instance y=4x-6 if you plug x=2 in, then you get y=2, which is the same answer we get when we plug x=2 into f(x)=x^2-2. So we have checked our slope with the power rule, and checked our tangent line by plugging in x=2, so we know it is correct.

To show understanding, tell me the slope of the tangent line of $f(x)=20x^2+3x+3.5$ at x=0.5? And then what is the tangent line? With the rule I showed you, you should be able to do this in just a few short steps.