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- Consider a differentiable function g with domain [0,infinity).Given that g'(x)=a-ex^2 with a an element of all real numbers and a>1The function has only one critical point at x2 =ln(a) .Use the Second Derivative test to identify the local extreme at x2 =ln(a) .Let f(x)=x+(1−x)^(1/2) Find the local maximum and minimum values of f using both the first and second derivative tests. How do I set this problem up and solve it?A function f has derivative f′(x)=x^3(x−1)^2(x+1)(x−2). At what numbers x, if any, does f have a local maximum? A local minimum?
- 1. Use the Cauchy-Riemann equation to determine if the function f(z) = x3 - i(2 - y)3 is analytic or not. Provide all the sufficient conditions and the domain of analyticity and then find the derivative if it exists.Consider the cubic polynomial f(x) = x^3 -x. Find the equation of the tangent line of f(x) at the point (1, 0). Plot both f(x) and the tangent line on the same set of axes (label both). On a new set of axes, plot both f(x) and f'(x) (label both graphs). What is noticeable between f(x) and f'(x) ?6 From algebraic form of function ez, prove that |ez| = ex
- Show that f is continuous on (−∞, ∞). f(x) = 1 − x2 if x ≤ 1 ln(x) if x > 1 On the interval (−∞, 1), f is ---Select--- a polynomial an exponential a root a logarithmic a rational function; therefore f is continuous on (−∞, 1). On the interval (1, ∞), f is ---Select--- a polynomial an exponential a root a logarithmic a rational function; therefore f is continuous on (1, ∞). At x = 1, lim x→1− f(x) = lim x→1− = , and lim x→1+ f(x) = lim x→1+ = , so lim x→1 f(x) = . Also, f(1) = . Thus, f is continuous at x = 1. We conclude that f is continuous on (−∞, ∞).Using the Forward Divided Difference Method, manually solve for the approximatevalue of the first derivative f’(xi) at xi = 0.5 of the given function, the absolute relative trueerror, εt, and absolute relative approximate error, εa, using step sizes ∆x = 0.5 and ∆x =0.25. Use five (5) decimal places in evaluating the first derivative of the function andcomputing for the errors.f(x) = 5x3 - cos(2x)Explain why Newton's method doesn't work for finding the root of the equation x3 − 3x + 3 = 0 if the initial approximation is chosen to be x1 = 1. f(x) = x3 − 3x + 3 ⇒ f '(x) = . If x1 = 1, then f '(x1) = and the tangent line used for approximating x2 is ---Select--- horizontal vertical . Attempting to find x2 results in trying to ---Select--- divide multiply by zero.
- How can I simplify f(x) = arctan(2(x-1)) - ln IxI such that I can describe the intervals of the function and calculate the approximations of the roots with bisection method or newton's method in python? I tried to put it into two functions f(x) = arctan(2x-2)-ln(x) and g(x) = arctan(2x-2)-ln(-x) as ln(x) and ln(-x) is the absolute value of ln(x) but it somehow gets too confusing when implementing it into bisection or newton's method. What can I do to simply this? Thank you very much for your help.1. Find the absolute maximum and minimum values of f(x)=10x(2-lnx) on [1,e^2] 2. Find the limil. Uisng L'Hopital's Rule : lim x arrow 0 sin^2(3x)/4xa. Locate the critical points of ƒ.b. Use the First Derivative Test to locate the local maximum and minimum values.c. Identify the absolute maximum and minimum values of the functionon the given interval (when they exist). ƒ(x) = √x ln x on (0, ∞)