Question

A curve passes through the point (0,18),(1,10),(3,-18) and (6,90) .Find the slope of the curve at x=2​

Answer 1

Answer:

100%

Inspection

500

20%

Administrative overhead

5,000

25%

Selling overhead

3,000

50%

Depreciation

10,000

100% fixed

Total

62,750

Cost

per

unit

12:55

You are required to prepare the production cost budget (flexible) at 4,000 units and 6,000 units.

TYY

Answer 2

Answer:

Slope of the curve at x=2 is -16

Explanation:

Given that,

A curve passes through the points (0,18),(1,10),(3,-18) and (6,90).We are required to find its slope at x=2.

To obtain slope,we need to find the equation of the given function.To find it,we shall use Lagrange`s Interpolation method.

According to that method,the function y=f(x) which passes through certain points is given as-

$f(x)=\frac{(x-x_{1})(x-x_{2})(x-x_{3})}{(x_{0}-x_{1})(x_{0}-x_{2})(x_{0}-x_{3})}y_{0} +\frac{(x-x_{0})(x-x_{2})(x-x_{3})}{(x_{1}-x_{0})(x_{1}-x_{2})(x_{1}-x_{3})}y_{1} +\frac{(x-x_{0})(x-x_{1})(x-x_{3})}{(x_{2}-x_{0})(x_{2}-x_{1})(x_{2}-x_{3})}y_{2} +\frac{(x-x_{0})(x-x_{1})(x-x_{2})}{(x_{3}-x_{0})(x_{3}-x_{1})(x_{3}-x_{2})}y_{3}$

For our given question,

$x_{0}=0 , y_{0}=18\\x_{1}=1 , y_{1}=10\\x_{2}=3 ,y_{2}=-18\\x_{3}=6 ,y_{3}=90$

Hence our function becomes

$f(x)=\frac{(x-1)(x-3)(x-6)}{-1.-3.-6} (18)+\frac{(x)(x-3)(x-6)}{1.-2.-5} (10)+\frac{(x)(x-1)(x-3)}{3.2.-3} (-18)+\frac{x(x-1)(x-6)}{6.5.3} (90)$On simplification of the above terms,the function is obtained as

$f(x)=2x^{3}-10x^{2}+18$

To find the slope of the curve at x=2,we shall differentiate it as shown below

$f`(x)=\frac{d}{dx} (2x^{3}-10x^{2}+18)=6x^{2}-20x$

Hence slope at given point is $f`(2)=6(2)^{2}-20(2)=24-40=-16$

Therefore,Slope of the curve at x=2 is -16

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