If f(x)=(2x-1)(3x+2)(4x-3) is a real-valued f | vedclass

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(D) Given $f(x) = (2x-1)(3x+2)(4x-3)$.First, check the conditions of Rolle's Theorem:$f(\frac{1}{2}) = (2(\frac{1}{2})-1)(...) = 0 	imes (...) = 0$.$

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$\frac{7+\sqrt{247}}{36}$

Vedclass · Answer

(D) Given $f(x) = (2x-1)(3x+2)(4x-3)$.First, check the conditions of Rolle's Theorem:$f(\frac{1}{2}) = (2(\frac{1}{2})-1)(...) = 0 	imes (...) = 0$.$f(\frac{3}{4}) = (...)(4(\frac{3}{4})-3) = (...)(3-3) = 0$.Since $f(\frac{1}{2}) = f(\frac{3}{4}) = 0$ and $f(x)$ is a polynomial, it is continuous and differentiable.Expanding $f(x)$:$f(x) = (6x^2 + 4x - 3x - 2)(4x-3) = (6x^2 + x - 2)(4x-3)$$f(x) = 24x^3 - 18x^2 + 4x^2 - 3x - 8x + 6 = 24x^3 - 14x^2 - 11x + 6$.Now, find $f'(x)$:$f'(x) = 72x^2 - 28x - 11$.Set $f'(c) = 0$:$72c^2 - 28c - 11 = 0$.Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$c = \frac{28 \pm \sqrt{(-28)^2 - 4(72)(-11)}}{2(72)} = \frac{28 \pm \sqrt{784 + 3168}}{144} = \frac{28 \pm \sqrt{3952}}{144}$.$c = \frac{28 \pm 4\sqrt{247}}{144} = \frac{7 \pm \sqrt{247}}{36}$.Since $\sqrt{247} \approx 15.7$, $c_1 = \frac{7+15.7}{36} \approx 0.63$ and $c_2 = \frac{7-15.7}{36} \approx -0.24$.Since $c \in [0.5, 0.75]$, only $c = \frac{7+\sqrt{247}}{36}$ is valid.

If $f(x)=(2x-1)(3x+2)(4x-3)$ is a real-valued function defined on $[\frac{1}{2}, \frac{3}{4}]$, then the value$(s)$ of '$c$' as defined in the statement of Rolle's theorem is/are:

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