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

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0). 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(c)=0 ). find the derivative of ( f(x)=x^{4}+5x + 3 ). ( f(x)=4x^{3}+5 ) can the derivative of ( f(x) ) be zero in the interval (-1,0)? yes no the function ( f(x)=x^{4}+5x + 3 ) has at least one zero at some point ( x = a ) in the interval (-1,0). according to rolles theorem, can there be another point ( x = b ) in this interval where ( f(a)=f(b)=0 )? yes no

show that the function ( f(x)=x^{4}+5x + 3 ) has exactly one zero in the interval (-1,0). 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(c)=0 ). find the derivative of ( f(x)=x^{4}+5x + 3 ). ( f(x)=4x^{3}+5 ) can the derivative of ( f(x) ) be zero in the interval (-1,0)? yes no the function ( f(x)=x^{4}+5x + 3 ) has at least one zero at some point ( x = a ) in the interval (-1,0). according to rolles theorem, can there be another point ( x = b ) in this interval where ( f(a)=f(b)=0 )? yes no

Answer

Explanation:

Step1: Analyze the derivative (f'(x)=4x^{3}+5) in the interval ([-1,0])

For (x\in[-1,0]), the function (y = 4x^{3}) is increasing (since the derivative of (y = 4x^{3}) with respect to (x) is (y'=12x^{2}\geq0)). When (x=-1), (4x^{3}+5=4\times(-1)^{3}+5=-4 + 5=1); when (x = 0), (4x^{3}+5=4\times0^{3}+5=5). So (f'(x)=4x^{3}+5>0) for all (x\in[-1,0]).

Step2: Apply the Intermediate Value Theorem

(f(-1)=(-1)^{4}+5\times(-1)+3=1 - 5+3=-1), (f(0)=0^{4}+5\times0 + 3=3). Since (f(x)) is continuous on ([-1,0]) (as a polynomial) and (f(-1)<0), (f(0)>0), by the Intermediate Value Theorem, there exists at least one (c\in(-1,0)) such that (f(c)=0).

Step3: Use the fact that (f(x)) is strictly increasing

Since (f'(x)>0) for all (x\in[-1,0]), (f(x)) is strictly increasing on ([-1,0]). If there were two zeros (a) and (b) ((a < b)) in ([-1,0]) such that (f(a)=f(b)=0), then by Rolle's Theorem, there would exist a (d\in(a,b)) such that (f'(d)=0). But we have shown (f'(x)>0) for all (x\in[-1,0]). So there can be at most one zero.

Combining the existence (from the Intermediate Value Theorem) and uniqueness (from the fact that (f(x)) is strictly increasing) of the zero, we conclude that (f(x)) has exactly one zero in ([-1,0]).

For the multiple - choice questions:

  • For the question "Can the derivative of (f(x)) be zero in the interval ([-1,0])": Since (f'(x)=4x^{3}+5), and for (x\in[-1,0]), (4x^{3}\in[-4,0]), so (f'(x)=4x^{3}+5\in[1,5]). So the answer is No.
  • For the question "According to Rolle's Theorem, can there be another point (x = b) in this interval where (f(a)=f(b)=0)": Since (f'(x)\neq0) for all (x\in[-1,0]), by Rolle's Theorem (if (f(a) = f(b)=0) for (a,b\in[-1,0]), then (f'(c)=0) for some (c\in(a,b))), the answer is No.

Answer:

For "Can the derivative of (f(x)) be zero in the interval ([-1,0])": No. For "According to Rolle's Theorem, can there be another point (x = b) in this interval where (f(a)=f(b)=0)": No.