Question

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0).
which theorem can be used to determine whether a function ( f(x) ) has any zeros in a given interval?
a. intermediate value theorem
b. extreme value theorem
c. rolles theorem
d. mean value theorem
to apply this theorem, evaluate the function ( f(x)=x^{4}+5x + 3 ) at each endpoint of the interval (-1,0).
( f(-1)=-1 ) (simplify your answer.)
( f(0)=3 ) (simplify your answer.)
according to the intermediate value theorem, ( f(x)=x^{4}+5x + 3 ) has in the given interval

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

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.