Question

Question a and b
I would really appreciate if you could help me with this question.

Answer 1

Part A)

`\displaystyle (dy)/(dx)=x^2+2x-7`

Part B)

`\displaystyle \Big(-4,(53)/(3)\Big)\text{ and } \Big(2, -(37)/(3)\Big)`

Explanation:

We are given the function:

`\displaystyle y=(x^3)/(3)+x^2-7x-5`

Part A)

To find dy/dx, differentiate both sides with respect to x:

`\displaystyle (dy)/(dx)=(d)/(dx)\Big[(x^3)/(3)+x^2-7x-5\Big]`

Differentiate:

`\displaystyle (dy)/(dx)=x^2+2x-7`

Part B)

We want the points on the curve where the gradient is parallel to y = x.

The equation y = x has a constant gradient of 1.

Therefore, we can set dy/dx = 1 and solve for x:

`1=x^2+2x-7`

Rewrite:

`x^2+2x-8=0`

Factor:

`(x+4)(x-2)=0`

Thus:

`x=-4\text{ and } x=2`

And substituting them back for the original equation, we acquire:

`\displaystyle y(-4)=((-4)^3)/(3)+(-4)^2-7(-4)-5=(53)/(3)`

And:

`\displaystyle y(2)=((2)^3)/(3)+(2)^2-7(2)-5=-(37)/(3)`

Our points are:

`\displaystyle \Big(-4,(53)/(3)\Big)\text{ and } \Big(2, -(37)/(3)\Big)`