Final answer:
The linearization of the function f(x) = x⁶ at a = 3 is L(x) = 729 + 1458(x - 3). The approximations for 8.98 and 9.02 using the linearization are both overestimates.
Step-by-step explanation:
The linearization of the function f(x) = x⁶ at a = 3 can be found by using the point-slope form of a linear equation. Firstly, we find the slope of the function at a = 3 by taking the derivative: f'(x) = 6x⁵. Substituting a = 3, we get the slope: f'(3) = 6(3)⁵ = 6(243) = 1458. Now, using the point-slope form, we have the linearization L(x) = f(a) + f'(a)(x-a). Substituting the values, we get L(x) = 729 + 1458(x - 3).
To approximate the numbers 8.98 and 9.02, we substitute these values into the linearization. For 8.98, we have L(8.98) = 729 + 1458(8.98 - 3) = 4745.64. For 9.02, we have L(9.02) = 729 + 1458(9.02 - 3) = 4753.64.
Comparing these approximations to the actual function values, we can see that both approximations are overestimates, as the linearization is slightly greater than the actual function values.