Show that the function flx)- x+8x+5 has exactly one zero in the interval [-1, 01. Which...
At least one of the answers above is NOT correct. (1 point) Suppose /(x) = x + 3x + 1. In this problem, we will show that has exactly one root (or zero) in the interval (-3,-1). (a) First, we show that f has a root in the interval (-3,-1). Since is a continuous function on the interval (-3, -1) and f(-3) = and f(-1) = -1 the graph of y = f(x) must cross the X-axis at some point...
Use the intermediate value theorem to show that the polynomial function has a zero in the given interval. f(x) = 2x® + 3x2 – 2x+8; (-8, -2] Find the value of f(-8). f(-8)= (Simplify your answer.) Find the value of f(-2). f(-2)= (Simplify your answer.) According to the intermediate value theorem, does f have a zero in the given interval? Yes Νο Ο
*5. A function f defined on an interval I = {x: a <x<b> is called increasing f(x) > f(x2) whenever xi > X2 where x1, x2 €1. Suppose that has the inter- mediate-value property: that is, for each number c between f(a) and f(b) there is an x, el such that f(x) = c. Show that a functionſ which is increasing and has the intermediate-value property must be continuous. This is from my real analysis textbook, We are establishing the...
The function f that satisfies f'(x) = 3x2 +1, f(2)=5 is Select one: a. f(2)=23+ O b. f(x) = 2 + 1 - 5 O c. f(x)=x3 ++2 d. f(x) = 3x2 – 7 O e. None of these. Let f(2) be a function that is continuous on 1, 7] with f(1) =3 and f(7) = 9. Then the Intermediate Value Theorem guarantees that Select one: O a. f'(2)=1 has at least on solution in the open interval (1,7). O...
2. The function (-3x if 0sx < 1 if x 1 -fO f(x) =f(x) 0 Is zero at x 0 and x = 1 and differentiable on (0,1), but its derivative on (0,1) is never zero. Does this example contradict Rolle's Theorem? Why or why not? 2. The function (-3x if 0sx
Use the intermediate value theorem to show that the polynomial has a real zero between the given integers. f(x) = 4x3 - 2x - 5; between 1 and 3 Select the correct choice below and, if necessary, fill in the answer box(es) within your choice. (Simplify your answers.) A. Because f(x) is a polynomial with f(1) = <0 and f(3) = <0, the function has a real zero between 1 and 3. B. Because f(x) is a polynomial with f(1)...
Let s < t and let f:[s,t]→ℝ be a differentiable function. Suppose that f'(x) > 0 for all x Which of the following is correct? 1. Using the definition of the derivative, it follows that f(x)<f(a) for any x<a. 2. Using Rolle's Theorem, it follows that f is a continuous function. 3. Using the Mean Value Theorem, it follows that f(t)>f(s). 4. Using Rolle's Theorem, it follows that there is some x∈[s,t] such that f′(x)=0. 5. Using the Intermediate Value...
Using the Intermediate Value Theorem explain why the function has at least one zero on the given interval, include as much detail as possible and work. Interval 87. Function h(x) = -2e-1/2 cos 22 T
Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. (Select all that apply.)f (x) = sin(x), [0, 2π]If Rolle's Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = 0. (Enter your answers as a comma-separated list. If Rolle's Theorem cannot be applied, enter NA.)I thought the derivative would be cos(x) so then cos(0) would be 1 but thatz wrong so now I don't understand...
a. Determine whether the Mean Value Theorem applies to the function f(x) = x + on the interval [3,6]. b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem. a. Choose the correct answer below. O O A. No, because the function is not continuous on the interval [3,6], and is not differentiable on the interval (3,6). B. No, because the function is differentiable on the interval (3,6), but is not continuous...