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STatiana [176]
4 years ago
7

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

ached] to get the answer -4/3 ?? Isn't 4/3 the minimum value???

Mathematics
1 answer:
aliina [53]4 years ago
4 0
y=x(x-2)^2
\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0
\implies x=\dfrac23,x=2

are the critical points, and judging by the picture alone, you must have b=\dfrac23 and a=2. (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of \dfrac{\mathrm dy}{\mathrm dx}=y', which entails first finding the second derivative:

y'=3x^2-8x+4
\implies y''=6x-8

setting equal to 0 and finding the critical point:

6x-8=0\implies x=\dfrac86=\dfrac43

This is to say the minimum value of \dfrac{\mathrm dy}{\mathrm dx} *occurs when x=\dfrac43*, but this is not necessarily the same as saying that \dfrac43 is the actual minimum value.

The minimum value of \dfrac{\mathrm dy}{\mathrm dx} is obtained by evaluating the derivative at this critical point:

m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43
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