f(x)=x^3-3x^2+3x+1 near 2 (at a=2)
A. Use Linear approximation to approximate.
b.is this underestimate or overestimate? Justify answer.
C.Approximate f(2.25) using the linear approximation and compare the answer with the exact value.
Thanks.
f'(x) = 3x^2 - 6x + 3
f''(x) = 6x - 6
f''(2) > 0, so the graph is concave up at x=2. So, a linear approximation will underestimate f(x)
f'(2) = 3
f(2) = 3
the line through (2,3) with slope 3 is
y = 3(x-2)+3= 3x-3
y(2.25) = 3.75
f(2.25) = 3.953
the line underestimates
To approximate the value of f(x) near a specific point 'a', we can use linear approximation, also known as tangent line approximation. The formula for linear approximation is given by:
L(x) = f(a) + f'(a)(x - a)
where f(a) is the value of the function at point 'a', f'(a) is the derivative of the function evaluated at point 'a'.
A. To approximate f(x) = x^3-3x^2+3x+1 near a=2, we need to calculate f(2), f'(2), and substitute these values into the formula.
1. Calculate f(2):
f(2) = (2)^3 - 3(2)^2 + 3(2) + 1 = 8 - 12 + 6 + 1 = 3
2. Calculate f'(2):
To find the derivative of f(x), we differentiate the function with respect to x:
f'(x) = 3x^2 - 6x + 3
Evaluate f'(2):
f'(2) = 3(2)^2 - 6(2) + 3 = 12 - 12 + 3 = 3
3. Substitute values into the linear approximation formula:
L(x) = f(2) + f'(2)(x - 2)
Substituting the values:
L(x) = 3 + 3(x - 2)
B. To determine whether the linear approximation underestimates or overestimates the actual function value, we need to analyze the sign of the linear term, f'(2)(x - 2). Since f'(2) = 3 is positive, the sign of the linear term will be the same as the sign of (x - 2).
Since the point of approximation is 2, any value of x greater than 2 will make (x - 2) positive, and any value of x less than 2 will make (x - 2) negative.
Since the linear term is positive, the linear approximation will be an overestimate of the actual function value at any x value greater than 2.
C. To approximate f(2.25) using the linear approximation, we substitute x = 2.25 into the linear approximation formula:
L(x) = 3 + 3(x - 2)
L(2.25) = 3 + 3(2.25 - 2)
L(2.25) = 3 + 3(0.25)
L(2.25) = 3 + 0.75
L(2.25) = 3.75
To compare the result with the exact value, we can calculate f(2.25) using the original function:
f(2.25) = (2.25)^3 - 3(2.25)^2 + 3(2.25) + 1
f(2.25) = 11.390625
Comparing the linear approximation L(2.25) with the exact value f(2.25), we see that L(2.25) = 3.75, whereas f(2.25) = 11.390625. Therefore, the linear approximation is not very accurate near x = 2.25.