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2 years ago

PLEASE HELLPO I NEED SOMEONE PLS

Mathematics

1 answer:

Digiron [165]2 years ago

7 0

Answer (if it is possible check the answer in other sources):

1: y=7x+5.553;

2: y=7x+7.99.

Step-by-step explanation:

all the details can be found in the attachment (the answer is marked with colour).

1) to determine f'(x);

2) to calculate the values x₀ within the given interval;

3) to calculate the values f(x₀);

4) to calculate the values of interception 'i';

5) to write the required equations using 's' and 'i'.

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3 years ago

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

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$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
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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:

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

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:

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4 years ago