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)=square ) (simplify your answer.)

Explanation:

Step1: Evaluate \( f(-1) \)

Substitute \( x = -1 \) into \( f(x)=x^{4}+5x + 3 \).

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Step2: Evaluate \( f(0) \)

Substitute \( x = 0 \) into \( f(x)=x^{4}+5x + 3 \).

$$ f(0)=0^{4}+5\times0 + 3=3 $$

Since \( f(x)=x^{4}+5x + 3 \) is a polynomial, it is continuous on \( [-1,0] \). By the Intermediate Value Theorem, since \( f(-1)=-1\) and \( f(0) = 3\), and \( -1<0<3 \), there exists at least one \( c\in(-1,0) \) such that \( f(c)=0 \).

Now, find the derivative \( f^{\prime}(x)=4x^{3}+5 \). For \( x\in[-1,0] \), \( f^{\prime}(x)=4x^{3}+5\). When \( x\in[-1,0] \), \(4x^{3}\in[-4,0]\), so \( f^{\prime}(x)=4x^{3}+5>0\) (because \(4x^{3}+5\geq - 4 + 5=1>0\) for \(x\in[-1,0]\)). A function with a positive - derivative on an interval is strictly increasing. A strictly increasing function can cross the \(x\) - axis at most once.

Answer:

\( f(-1)=-1 \)