determine whether the mean value theorem can be applied to ( f ) on the closed interval ( a, b ). (select…

determine whether the mean value theorem can be applied to ( f ) on the closed interval ( a, b ). (select all that apply.)\n( f(x)=x^{3}+4 x + 3, quad-1,0 )\nyes, the mean value theorem can be applied.\nno, ( f ) is not continuous on ( a, b ).\nno, ( f ) is not differentiable on ( (a, b) ).\nnone of the above.\nif the mean value theorem can be applied, find all values of ( c ) in the open interval ( (a, b) ) such that ( f^{prime}(c)=\frac{f(b)-f(a)}{b - a} ). (enter your answers as a comma - separated list. if the mean value theorem cannot be applied, enter na.)\n( c = )

determine whether the mean value theorem can be applied to ( f ) on the closed interval ( a, b ). (select all that apply.)\n( f(x)=x^{3}+4 x + 3, quad-1,0 )\nyes, the mean value theorem can be applied.\nno, ( f ) is not continuous on ( a, b ).\nno, ( f ) is not differentiable on ( (a, b) ).\nnone of the above.\nif the mean value theorem can be applied, find all values of ( c ) in the open interval ( (a, b) ) such that ( f^{prime}(c)=\frac{f(b)-f(a)}{b - a} ). (enter your answers as a comma - separated list. if the mean value theorem cannot be applied, enter na.)\n( c = )

Answer

Explanation:

Step1: Check continuity and differentiability

A polynomial function (y = f(x)=x^{3}+4x + 3) is continuous and differentiable for all real (x). Since ([-1,0]) is a sub - interval of ((-\infty,\infty)), (f(x)) is continuous on ([-1,0]) and differentiable on ((-1,0)). So, the Mean Value Theorem can be applied.

Step2: Calculate (f(a)) and (f(b))

For (a=-1) and (b = 0): (f(-1)=(-1)^{3}+4(-1)+3=-1 - 4 + 3=-2) (f(0)=(0)^{3}+4(0)+3=3)

Step3: Calculate (\frac{f(b)-f(a)}{b - a})

(\frac{f(0)-f(-1)}{0-(-1)}=\frac{3-(-2)}{1}=5)

Step4: Find (f^{\prime}(x))

Differentiate (f(x)=x^{3}+4x + 3) using the power rule ((x^{n})^\prime=nx^{n - 1}). Then (f^{\prime}(x)=3x^{2}+4)

Step5: Solve (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a})

Set (f^{\prime}(c)=3c^{2}+4) equal to (5) (from Step 3). [ \begin{align*} 3c^{2}+4&=5\ 3c^{2}&=1\ c^{2}&=\frac{1}{3}\ c&=\pm\frac{1}{\sqrt{3}} \end{align*} ] Since (c\in(-1,0)), (c =-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3})

Answer:

Yes, the Mean Value theorem can be applied. (c=-\frac{\sqrt{3}}{3})