Let f(x) = x^2 -2x
a) Estimate f'(3) using h = 0.03
b) Find the exact value for f'(3)
c) Give the equation of the tangent line to f(x) at x=3
d) Use the tangent line to estimate f(2.98)
e) Critical points
(a)
(f(3+.03)-f(3))/.03 = (3.1209-3)/.03 = 4.03
(b)
f'(x) = 2x-2
f'(3) = 4
(c)
f(3) = 3
So, now you have a point and a slope. The line is easy to express.
(d)
Now just plug 2.98 into that line.
(e)
where is f'(x) = 0?
Thanks!
To answer these questions, we first need to find the derivative of the function f(x) = x^2 - 2x. Let's start by finding f'(x).
a) To estimate f'(3) using h = 0.03, we can use the formula for the derivative:
f'(x) = lim(h->0) [f(x+h) - f(x)]/h
Plugging in the values, we have:
f'(3) ≈ [f(3 + 0.03) - f(3)] / 0.03
To find f(3 + 0.03), substitute x = 3 + 0.03 into the original function:
f(3 + 0.03) = (3 + 0.03)^2 - 2(3 + 0.03)
Evaluating this expression, we find:
f(3 + 0.03) ≈ 8.4191
Now, substitute x = 3 into the original function to find f(3):
f(3) = 3^2 - 2(3)
Evaluating this, we find:
f(3) = 3
Now, we can calculate the estimated value for f'(3):
f'(3) ≈ (8.4191 - 3) / 0.03
Simplifying, we get:
f'(3) ≈ 137.303
Therefore, the approximate value of f'(3) using h = 0.03 is 137.303.
b) To find the exact value for f'(3), we can take the derivative of the function f(x) = x^2 - 2x.
Differentiating f(x) with respect to x, we get:
f'(x) = 2x - 2
Substitute x = 3 into the equation:
f'(3) = 2(3) - 2
Calculating this, we find:
f'(3) = 4
Hence, the exact value for f'(3) is 4.
c) To find the equation of the tangent line to f(x) at x = 3, we need two pieces of information: the slope of the tangent line (which is equal to the derivative at x = 3) and a point on the line (which is the point (3, f(3)).
We already found that f'(3) = 4. The point on the line is (3, f(3)). Evaluating f(3):
f(3) = (3)^2 - 2(3)
Simplifying, we get:
f(3) = 3
Therefore, the point on the line is (3, 3).
Using the slope-point form of a line, we can write the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values, we have:
y - 3 = 4(x - 3)
Expanding and rearranging, we get:
y = 4x - 9
Therefore, the equation of the tangent line to f(x) at x = 3 is y = 4x - 9.
d) To use the tangent line to estimate f(2.98), we need to substitute x = 2.98 into the equation of the tangent line:
y = 4(2.98) - 9
Evaluating this expression, we get:
y ≈ 3.92
Therefore, the estimate for f(2.98) using the tangent line is 3.92.
e) Critical points are the points where the derivative is either zero or undefined.
To find the critical points, we need to find the values of x where f'(x) = 0 or where f'(x) is undefined.
We already found that f'(x) = 2x - 2. To find the critical points, we set the derivative equal to zero and solve for x:
2x - 2 = 0
Adding 2 to both sides, we get:
2x = 2
Dividing by 2, we find:
x = 1
Hence, the critical point is x = 1.
Since the derivative is defined for all real numbers, there are no points where the derivative is undefined in this case.
Therefore, the critical point for the function f(x) = x^2 - 2x is x = 1.