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)=square ) (simplify your answer.)

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)=square ) (simplify your answer.)

Answer

Explanation:

Step1: Evaluate ( f(-1) )

Substitute ( x = -1 ) into ( f(x)=x^{4}+5x + 3 ). [ \begin{align*} f(-1)&=(-1)^{4}+5\times(-1)+3\ &=1 - 5+3\ &=-1 \end{align*} ]

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 )