Find the exact coordinates of the inflection points and critical points of the function?

Question

f(x)=2x^3+3x^2-180+150 on the interval (-10, 10)

Answer 1

4

Answer 2

f(x) = 2x^3 + 3x^2 - 180x + 150 on the interval (-10, 10)

(I assume you forgot an x in front of the 180).

To solve for the critical points, you have to take the derivative and make it 0.

f'(x) = 6x^2 + 6x - 180. Now, making f'(x) = 0,
0 = 6x^2 + 6x - 180. Dividing everything by 6,
0 = x^2 + x - 30

Now, we can factor.

0 = (x + 6) (x - 5); therefore x = {-6, 5}
Both of these values lie on our interval (-10, 10).

Our critical points are x = -6 and x = 5.
To obtain the coordinates of our critical points, we just plug in these values for our function.

f(-6) = 2(-6)^3 + 3(-6)^2 - 180(-6) + 150

I'm not going to show you the algebra in detail. Your final answer should be
f(-6) = 474
f(5) = 2(5)^3 + 3(5)^2 - 180(5) + 150 = -425

Coordinates of critical points: (-6, 474) (5, -425)

To find the inflection points, we take the second derivative and make it 0. Since

f'(x) = 6x^2 + 6x - 180, then
f''(x) = 12x + 6. Making f''(x) = 0, we get
0 = 12x + 6
-12x = 6
x = -1/2 (which is also in our interval (-10, 10)

To solve for the coordinates of the inflection point, plug in f(-1/2)

I'll leave it up to you to finish.

Answer 3

Ok...did you mean 180*x? If you did then there is a local max at (-6,906) and a local min at (5,-425) and an inflection pt at (-1/2,481/2).......

> f1:=diff(2*x^3+3*x^2-180*x+150,x);
> solve(f1=0,x);
2
f1 := 6 x + 6 x - 180
-6, 5
> subs(x=-6,2*x^3+3*x^2-180*x+150);
> subs(x=5,2*x^3+3*x^2-180*x+150);
906
-425
> f2:=diff(f1,x);
> solve(f2=0,x);
f2 := 12 x + 6
-1/2
> subs(x=-1/2,2*x^3+3*x^2-180*x+150);
481/2

IF you didn't mean 180x but really meant 180 then the critical pts are at (-1,-29) and (0,-30) and inflection pt at (-1/2, -59/2)