Question

2. Find the stationary points for the function

Answer 1

Answer & step-by-step explanation:

Stationary points are the points where the first derivative is equal to zero.

Let's calculate it using the power rule (exponent comes forward, decrease exponent by 1) and the fact that the derivative is a linear operation (that is $D[a\ f(x) + b\ g(x)] = a Df(x) + b Dg(x)$)

The first derivative is then

$y' = \frac13 (3x^2) - \frac12 (2x) -6 = x^2-x-6 = (x+2)(x-3)$

Note that the last passage is not strictly needed, but it's really helpful to find stationary points, when in this next passage we set it equal to zero. Alternatively, you can use the quadratic formula if you can pull the factors out of your head right away.

$y'=0 \rightarrow (x+2)(x-3) = 0 \implies x=-2 || x=3$

These two point could be maxima, minima, or inflection points. To check them you can either see how the sign of the first derivative goes, or check the sign of the second derivative, as you're required.

The rules states that if the second derivative evaluated in that point is negative we have a maximum, if it's positive we have a minimum, and if we have a zero we keep derivating until we get a non-zero  value.

In our case, the second derivative we get by calculating the derivative again and we get $y'' = 2x-1$ . Evaluating it at both points we get

$y''(-2) = 2(-2)-1 = -5\\y''(3) = 2(3) -1 = 5$

so -2 is a maximum and 3 is a minimum.

Answer 2

Answer:

$\left( 3, -\dfrac{13}{2} \right) \textsf{minimum}$

$\left( -2, -\dfrac{43}{3} \right)\textsf{maximum}$

Step-by-step explanation:

To find the stationary points, differentiate, set to zero, and solve for x.  Input found values of x into original equation to find y-coordinates.

$y=\dfrac13x^3-\dfrac12x^2-6x+7$

$\implies \dfrac{dy}{dx}=3\cdot \dfrac13x^{(3-1)}-2 \cdot \dfrac12x^{(2-1)}-6x^{(1-1)}+7\cdot0$

$\implies \dfrac{dy}{dx}=x^2-x-6$

$\dfrac{dy}{dx}=0$

$\implies x^2-x-6=0$

$\implies (x-3)(x+2)=0$

$\implies x=3, x=-2$

$\textsf{at }x=3: y=\dfrac13(3)^3-\dfrac12(3)^2-6(3)+7=-\dfrac{13}{2}$

$\textsf{at }x=-2: y=\dfrac13(-2)^3-\dfrac12(-2)^2-6(-2)+7=\dfrac{43}{3}$

To determine the nature of each point, differentiate again, input found values of x into the second derivative. If positive then minimum point, if negative then maximum point.

$\dfrac{d^2y}{dx^2}=2 \cdot x^{(2-1)}-x^{(1-1)}-6 \cdot0$

$\implies \dfrac{d^2y}{dx^2}=2x-1$

$\textsf{at }x=3:\dfrac{d^2y}{dx^2}=2(3)-1=5 > 0\implies \textsf{minimum}$

$\textsf{at }x=-2:\dfrac{d^2y}{dx^2}=2(-2)-1=-5 < 0\implies \textsf{maximum}$