show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0).\nwhich theorem can…

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0).\nwhich theorem can be used to determine whether a function ( f(x) ) has any zeros in a given interval?\na. intermediate value theorem\nb. extreme value theorem\nc. rolles theorem\nd. mean value theorem\nto apply this theorem, evaluate the function ( f(x)=x^{4}+5x + 3 ) at each endpoint of the interval (-1,0).\n( f(-1)=-1 ) (simplify your answer.)\n( f(0)=3 ) (simplify your answer.)\naccording to the intermediate value theorem, ( f(x)=x^{4}+5x + 3 ) has in the given interval

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0).\nwhich theorem can be used to determine whether a function ( f(x) ) has any zeros in a given interval?\na. intermediate value theorem\nb. extreme value theorem\nc. rolles theorem\nd. mean value theorem\nto apply this theorem, evaluate the function ( f(x)=x^{4}+5x + 3 ) at each endpoint of the interval (-1,0).\n( f(-1)=-1 ) (simplify your answer.)\n( f(0)=3 ) (simplify your answer.)\naccording to the intermediate value theorem, ( f(x)=x^{4}+5x + 3 ) has in the given interval

Answer

Explanation:

Step1: Analyze the function values at endpoints

The Intermediate Value Theorem states that if a function (y = f(x)) is continuous on a closed interval ([a,b]), and (k) is a number between (f(a)) and (f(b)), then there exists at least one number (c\in(a,b)) such that (f(c)=k).

For the function (f(x)=x^{4}+5x + 3), which is a polynomial (and polynomials are continuous everywhere, so continuous on ([-1,0])).

We have (f(-1)=(-1)^{4}+5\times(-1)+3=1 - 5+3=-1) and (f(0)=0^{4}+5\times0 + 3=3).

Since (0) is between (f(-1)=-1) and (f(0) = 3) (i.e., (-1<0<3)), by the Intermediate Value Theorem, there exists at least one (c\in(-1,0)) such that (f(c)=0).

Step2: Show the function is monotonic (to prove uniqueness)

Find the derivative of (f(x)) using the power rule. If (f(x)=x^{4}+5x + 3), then (f^\prime(x)=4x^{3}+5).

For (x\in[-1,0]):

When (x=-1), (f^\prime(-1)=4\times(-1)^{3}+5=-4 + 5=1>0)

When (x = 0), (f^\prime(0)=4\times0^{3}+5=5>0)

Since (y = 4x^{3}+5) is a continuous function (as it is a polynomial) and (f^\prime(x)=4x^{3}+5>0) for all (x\in[-1,0]) (because (y = 4x^{3}) is an increasing - function on (\mathbb{R}) and (4x^{3}\geq-4) for (x\in[-1,0]), so (4x^{3}+5\geq1>0)), the function (y = f(x)) is strictly increasing on the interval ([-1,0]).

A strictly increasing function (y = f(x)) (if (x_1<x_2), then (f(x_1)<f(x_2))) can cross the (x -)axis at most once.

Answer:

The function (f(x)=x^{4}+5x + 3) has exactly one zero in the interval ([-1,0]) because:

  1. By the Intermediate Value Theorem (since (f(-1)=-1), (f(0)=3) and (f(x)) is continuous on ([-1,0])), there is at least one zero.
  2. Since (f^\prime(x)=4x^{3}+5>0) for all (x\in[-1,0]), (f(x)) is strictly increasing on ([-1,0]), so there is at most one zero.