Question

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

Answer 1

Answer:

Part A)

$\displaystyle \frac{dy}{dx}=x^2+2x-7$

Part B)

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

Step-by-step explanation:

We are given the function:

$\displaystyle y=\frac{x^3}{3}+x^2-7x-5$

Part A)

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

$\displaystyle \frac{dy}{dx}=\frac{d}{dx}\Big[\frac{x^3}{3}+x^2-7x-5\Big]$

Differentiate:

$\displaystyle \frac{dy}{dx}=x^2+2x-7$

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:

$\displaystyle y(-4)=\frac{(-4)^3}{3}+(-4)^2-7(-4)-5=\frac{53}{3}$

And:

$\displaystyle y(2)=\frac{(2)^3}{3}+(2)^2-7(2)-5=-\frac{37}{3}$

Our points are:

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