Question

f(0)=3 (simplify your answer.) according to the intermediate value theorem, f(x) = x⁴ + 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(c)=0. find the derivative of f(x)=x⁴ + 5x + 3. f(x)=4x³ + 5 can the derivative of f(x) be zero in the interval -1, 0? yes no

Explanation:

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

For \(x\in[-1,0]\), when \(x=-1\), \(f^{\prime}(-1)=4\times(-1)^{3}+5=-4 + 5=1>0\). When \(x = 0\), \(f^{\prime}(0)=4\times0^{3}+5=5>0\). Also, the function \(y = f^{\prime}(x)=4x^{3}+5\) is a continuous function (since it is a polynomial). The derivative \(y = 4x^{3}+5\) is an increasing - function (because \(y^{\prime}=12x^{2}\geq0\) for all \(x\in R\)).

Step2: Determine if \(f^{\prime}(x)=0\) in \([-1,0]\)

Since \(f^{\prime}(x)=4x^{3}+5>0\) for all \(x\in[-1,0]\) (because \(x^{3}\in[-1,0]\) in the interval \([-1,0]\), so \(4x^{3}\in[-4,0]\) and \(4x^{3}+5\in[1,5]\)).

Answer:

B. No