Question

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)= )
a 0
b 9
c 10
d 14
e 16

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 \(\frac{f(3)-f(1)}{3 - 1}\)

$$
LATEXBLOCK0
$$

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