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

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0).\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 at least one zero in the given interval.\nnow, determine whether there can be more than one zero in the given interval.\nrolles 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 ).\nfind the derivative of ( f(x)=x^{4}+5x + 3 ).\n( f^{prime}(x)=square )

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0).\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 at least one zero in the given interval.\nnow, determine whether there can be more than one zero in the given interval.\nrolles 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 ).\nfind the derivative of ( f(x)=x^{4}+5x + 3 ).\n( f^{prime}(x)=square )

Answer

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)