this question concerns the function

Question

f(x)=-2x3+3x2+12x+10

(a)find the stationary points of this function

using the strategy to apply the first derivative test,classify the left hand stationary point found in part (a)

find the y-coordinate of each of the stationary points on the graph of the function f(x), and also evaluate f(0)

Answer 1

To find the stationary points of a function, we first need to find the points where the first derivative is equal to zero. So let's compute the derivative of the given function:

f(x) = -2x^3 + 3x^2 + 12x + 10

f'(x) = d/dx(-2x^3) + d/dx(3x^2) + d/dx(12x) + d/dx(10)
= -6x^2 + 6x + 12

Now let's set f'(x) = 0 and solve for x:

-6x^2 + 6x + 12 = 0
=> x^2 - x - 2 = 0

Now we factor the quadratic equation:

(x - 2)(x + 1) = 0

So, we get two possible stationary points, x1 = 2 and x2 = -1.

Now to classify the left-hand stationary point using the first derivative test, we need to check the sign of the derivative around x1 = -1:

f'(-1.1) = 6.61 - 6.6 + 12 > 0 (It is positive)
f'(-0.9) = 5.61 - 5.4 + 12 > 0 (It is positive)

Since f'(-1.1) > 0 and f'(-0.9) > 0, we can conclude that the left-hand stationary point, x1 = -1, is a local minimum.

Now, we need the y-coordinate for both of the stationary points:

For x1 = -1:
f(-1) = -2(-1)^3 + 3(-1)^2 + 12(-1) + 10 = -2 + 3 - 12 + 10 = -1

For x2 = 2:
f(2) = -2(2)^3 + 3(2)^2 + 12(2) + 10 = -16 + 12 + 24 + 10 = 30

The stationary points on the graph of the function are (-1, -1) and (2, 30).

Now we need to evaluate f(0):
f(0) = -2(0)^3 + 3(0)^2 + 12(0) + 10 = 0 + 0 + 0 + 10 = 10

So, f(0) = 10.