Question

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0).
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 at least one zero in the given interval.
now, determine whether there can be more than one zero in the given interval.
rolles theorem states that for a function ( f(x) ) that is continuous at every point over the closed interval (a,b) and differentiable at every point of its interior ( (a,b) ), if ( f(a)=f(b) ), then there is at least one number ( c ) in ( (a,b) ) at which ( f^{prime}(c)=0 ).
find the derivative of ( f(x)=x^{4}+5x + 3 ).
( f^{prime}(x)=square )

Explanation:

Step1: Differentiate term - by - term

Differentiate \(x^{4}\) using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(5x\) using the rule \((ax)^\prime=a\) and the constant \(3\) (since \((c)^\prime = 0\) for a constant \(c\)).
For \(y = x^{4}+5x + 3\), by the sum rule \((u + v+w)^\prime=u^\prime+v^\prime + w^\prime\) where \(u=x^{4}\), \(v = 5x\), \(w = 3\).
\(u^\prime=(x^{4})^\prime=4x^{3}\), \(v^\prime=(5x)^\prime=5\), \(w^\prime=(3)^\prime = 0\).

Step2: Combine the derivatives

\(f^\prime(x)=4x^{3}+5\)

Answer:

\(4x^{3}+5\)