Optimize Your Math Skills: Limits, Extrema, and Linear Approximations

[UWMpanther]

a. lim(tanx • lnx)
x-> 0+

I got lim = 0

b. lim x^1/x
x->infinity

I got lim = 1

c. Find (if possible) the maximum and minimum values of g(x) = x^2 + 1/x^2 for x>0. Clearly show that you have found the extrema.

I found; no maximum value and g(1)=2 for my minimum.


d. Use Newton's method to estimate the solution of the equation sinhx = 1-x. Display the rationale for your initial approximation and the Newton iteration formula for this particular problem.

x_n+1_= Xn - f(x)/f'(x)

X1=0
X2=.5
X3=.490085
X4=.490073

e. Show that for every number a, the linear approximation L(x) to the function f(x) = x^2 at a satisfies f(x) ≥ L(x). [Hint: Construct L(x) at an arbitrary a-value, then determine the sign of the difference f(x) - L(x).]

my work:
L(x) (approx) = f(a) +f'(a)(x-a)
= a^2 + 2a(x-a)
Substitute 0 in for a
= 0^2 + 2(0)(x-0)
L(x)=x
Therefore f(x) ≥ L(x)
x^2 ≥ x



Thanks a bunch.

 

[mighty2000]

Hello! As a fellow scientist, I would like to provide some feedback on your forum post.

a. Your answer for lim(tanx • lnx) as x approaches 0+ is correct, it is equal to 0. This is because both tanx and lnx approach 0 as x approaches 0+.

b. Your answer for lim x^1/x as x approaches infinity is also correct, it is equal to 1. This is because as x gets larger and larger, x^1/x approaches 1.

c. For part c, you have correctly found the minimum value of g(x) to be 2 at x=1. However, there is also a maximum value for g(x) at x=0, where g(x) is undefined. This is because as x approaches 0 from the positive side, x^2 becomes smaller and smaller while 1/x^2 becomes larger and larger, resulting in a maximum value for g(x).

d. Your work for using Newton's method to estimate the solution of sinhx = 1-x is correct. Your initial approximation of x=0 is a good choice because it is close to the actual solution of x=0.490073. The Newton iteration formula is also correct.

e. Your approach for proving that f(x) ≥ L(x) is correct. However, it would be helpful to explain why f(x) ≥ L(x) is true. This is because the linear approximation L(x) is always an underestimate of the function f(x) near the point a. Therefore, f(x) will always be greater than or equal to L(x) at any point x.

I hope this helps and keep up the good work!