y = 2x^3 - 8x^2 + 10x - 4 is a function defin | vedclass

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(D) Given the function $y = 2x^3 - 8x^2 + 10x - 4$. To find the slope of the tangent, we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 6x^2

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(D) Given the function $y = 2x^3 - 8x^2 + 10x - 4$. To find the slope of the tangent, we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 6x^2 - 16x + 10$. Since the tangent is parallel to the $X$-axis, its slope must be zero: $\frac{dy}{dx} = 0 \implies 6x^2 - 16x + 10 = 0$. Dividing by $2$, we get $3x^2 - 8x + 5 = 0$. Factoring the quadratic equation: $3x^2 - 3x - 5x + 5 = 0 \implies 3x(x - 1) - 5(x - 1) = 0$. $(3x - 5)(x - 1) = 0$. This gives $x = 1$ or $x = \frac{5}{3}$. Since the point $(a, b)$ satisfies $a \in (1, 2)$, we discard $x = 1$ and accept $a = \frac{5}{3}$.

$y = 2x^3 - 8x^2 + 10x - 4$ is a function defined on $[1, 2]$. If the tangent drawn at a point $(a, b)$ on the graph of this function is parallel to the $X$-axis and $a \in (1, 2)$, then $a =$

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