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(C) Given $f(x) = x^3 + ax^2 + bx + c$,where $a = f^{\prime}(1)$,$b = f^{\prime \prime}(2)$,and $c = -f^{\prime \prime \prime}(3)$.First,find the deri

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$45$

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(C) Given $f(x) = x^3 + ax^2 + bx + c$,where $a = f^{\prime}(1)$,$b = f^{\prime \prime}(2)$,and $c = -f^{\prime \prime \prime}(3)$.First,find the derivatives: $f^{\prime}(x) = 3x^2 + 2ax + b$,$f^{\prime \prime}(x) = 6x + 2a$,$f^{\prime \prime \prime}(x) = 6$.Now,solve for constants:$f^{\prime \prime \prime}(3) = 6$,so $c = -6$.$f^{\prime \prime}(2) = 6(2) + 2a = 12 + 2a$. Since $b = f^{\prime \prime}(2)$,we have $b = 12 + 2a$.$f^{\prime}(1) = 3(1)^2 + 2a(1) + b = 3 + 2a + b$. Since $a = f^{\prime}(1)$,we have $a = 3 + 2a + b$.Substitute $b = 12 + 2a$ into the equation for $a$: $a = 3 + 2a + (12 + 2a) \Rightarrow a = 15 + 4a \Rightarrow -3a = 15 \Rightarrow a = -5$.Then $b = 12 + 2(-5) = 2$.So,$f(x) = x^3 - 5x^2 + 2x - 6$.At $x=0$,$y = f(0) = -6$. The point is $(0, -6)$.Slope $m = f^{\prime}(0) = 3(0)^2 - 10(0) + 2 = 2$.Tangent at $(0, -6)$ is $y - (-6) = 2(x - 0) \Rightarrow y = 2x - 6$. $X$-intercept is $3$.Normal at $(0, -6)$ is $y - (-6) = -\frac{1}{2}(x - 0) \Rightarrow y = -\frac{1}{2}x - 6$. $X$-intercept is $-12$.The triangle is formed by the points $(0, -6)$,$(3, 0)$,and $(-12, 0)$.Base length $= 3 - (-12) = 15$. Height $= |-6| = 6$.Area $= \frac{1}{2} 	imes 15 	imes 6 = 45$ sq. units.

If $f: R \rightarrow R$ is a function defined for all $x \in R$ by $f(x)=x^3+f^{\prime}(1) x^2+f^{\prime \prime}(2) x-f^{\prime \prime \prime}(3)$,then the area (in sq. units) of the triangle formed by the $X$-axis,the tangent,and the normal drawn to the curve $y=f(x)$ at $x=0$ is

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