Would you be able to help me with a couple of questions? thanks in advanced!
1. I am given the graph f(x)=x^2-2x. I have to draw a line tangent to the point (0,0). Then, I have to estimate the slope at that point.
how would I go about answering this question? would I first find the slope? also how do you estimate the slope at that point?
2. Use the alternate definition, lim x->a f(x)-f(a)/x-a, to determine the instantaneous rate of change of f(x)= 1/x at the point (1,1/5)
For this question, I got b but I am not sure it is correct
you know the slope of the tangent for
y = f(x) at any point is
m = dy/dx
So, for y = x^2-2x, the slope at any x is
y'(x) = 2x - 2
So, at (0,0), mm = -2
If you want to estimate the slope, pick two x values close to zero and find the slope between those two points on the curve.
For example, if you pick -.1 and +.1, then
f(.1) = -.19
f(-.1) = .21
and the slope between (-.1,.21) and (+.1,-.19) is -.4/.2 = -2
No idea what "b" means for #2.
Of course, I'll be happy to help you with your questions!
1. To draw a line tangent to the point (0,0) on the graph of f(x) = x^2 - 2x, you first need to find the slope at that point. The slope of a tangent line is equal to the derivative of the function at that point. To find the derivative of f(x), you can use the power rule for differentiation.
f'(x) = 2x - 2.
Now, to find the slope at the point (0,0), substitute x = 0 into the derivative equation:
f'(0) = 2(0) - 2 = -2.
So, the slope of the tangent line at the point (0,0) is -2. To draw the line tangent to this point, you can use point-slope form. The equation of the line is:
y - 0 = -2(x - 0),
y = -2x.
2. To find the instantaneous rate of change or the slope of the tangent line at the point (1, 1/5) for f(x) = 1/x, you can use the alternate definition of the derivative:
lim x->a (f(x) - f(a))/(x - a).
In this case, a = 1, and f(x) = 1/x. Let's plug in the values:
lim x->1 (1/x - 1/(1))/(x - 1).
To simplify this expression, we need to find a common denominator:
lim x->1 [(1 - x)/(x*1)]/(x - 1),
lim x->1 (1 - x)/(x^2 - x).
The numerator simplifies to -(x - 1) and the denominator can be factored:
lim x->1 -(x - 1)/(x(x - 1)).
Cancel out the common factor of (x - 1):
lim x->1 -1/x,
-1/1 = -1.
Therefore, the instantaneous rate of change or the slope of the tangent line at the point (1, 1/5) is -1.
I hope these explanations are helpful! Let me know if there's anything else I can assist you with.