the function f is defined by ( f(x)=2x^{3}-4x^{2}+1 ). the application of the mean value theorem to f on the…

the function f is defined by ( f(x)=2x^{3}-4x^{2}+1 ). the application of the mean value theorem to f on the interval ( 1leq xleq3 ) guarantees the existence of a value c, where ( 1lt clt3 ), such that ( f(c)= ) \na 0 \nb 9 \nc 10 \nd 14 \ne 16

the function f is defined by ( f(x)=2x^{3}-4x^{2}+1 ). the application of the mean value theorem to f on the interval ( 1leq xleq3 ) guarantees the existence of a value c, where ( 1lt clt3 ), such that ( f(c)= ) \na 0 \nb 9 \nc 10 \nd 14 \ne 16

Answer

Explanation:

Step1: Recall the Mean Value Theorem formula

The Mean Value Theorem states that if (y = f(x)) is continuous on ([a,b]) and differentiable on ((a,b)), then (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}), where (a = 1), (b = 3).

Step2: Calculate (f(1)) and (f(3))

  • For (x = 1): (f(1)=2\times1^{3}-4\times1^{2}+1=2 - 4+1=-1).
  • For (x = 3): (f(3)=2\times3^{3}-4\times3^{2}+1=2\times27-4\times9 + 1=54-36 + 1=19).

Step3: Calculate (\frac{f(3)-f(1)}{3 - 1})

[ \begin{align*} \frac{f(3)-f(1)}{3 - 1}&=\frac{19-(-1)}{3 - 1}\ &=\frac{19 + 1}{2}\ &=\frac{20}{2}\ &=10 \end{align*} ]

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

Differentiate (f(x)=2x^{3}-4x^{2}+1) using the power rule ((x^{n})^\prime=nx^{n - 1}). So (f^{\prime}(x)=6x^{2}-8x). But we don't need to solve (f^{\prime}(c) = 10) for (c) as per the Mean Value Theorem formula (\frac{f(b)-f(a)}{b - a}) gives the value of (f^{\prime}(c)).

Answer:

C. 10