Suppose f(x)=x(x+3)(x-2),where x in [-1,4]. T | vedclass

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(A) Given $f(x) = x(x+3)(x-2)$.Expanding the expression: $f(x) = x(x^2 + x - 6) = x^3 + x^2 - 6x$.Differentiating with respect to $x$: $f^{\prime}(x)

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$2$

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(A) Given $f(x) = x(x+3)(x-2)$.Expanding the expression: $f(x) = x(x^2 + x - 6) = x^3 + x^2 - 6x$.Differentiating with respect to $x$: $f^{\prime}(x) = 3x^2 + 2x - 6$.We are given $f^{\prime}(c) = 10$,so $3c^2 + 2c - 6 = 10$.Rearranging the equation: $3c^2 + 2c - 16 = 0$.Factoring the quadratic equation: $3c^2 + 8c - 6c - 16 = 0$.$c(3c + 8) - 2(3c + 8) = 0$.$(3c + 8)(c - 2) = 0$.Thus,$c = -\frac{8}{3}$ or $c = 2$.Since we require $c \in (-1, 4)$,we reject $c = -\frac{8}{3}$ as it lies outside the interval.Therefore,$c = 2$ is the correct value.

Suppose $f(x)=x(x+3)(x-2)$,where $x \in [-1,4]$. Then,a value of $c$ in $(-1,4)$ satisfying $f^{\prime}(c)=10$ is

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