Question

determine whether the mean value theorem can be applied to ( f ) on the closed interval ( a, b ). (select all that apply.)
( f(x)=x^{3}+4 x + 3, quad-1,0 )
yes, the mean value theorem can be applied.
no, ( f ) is not continuous on ( a, b ).
no, ( f ) is not differentiable on ( (a, b) ).
none of the above.
if 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.)
( c = )

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).

$$
LATEXBLOCK0
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Since \(c\in(-1,0)\), \(c =-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}\)

Answer:

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