What are all the values of x for the graph of f(x) = (x + 5)(x - 2)^2 ...and is concave downward?

Question

4/3 < x < 6 ??

Answer 1

Your equation works out to be a cubic function, i.e. it would have at most 2 maxima or minima (humps in the curve).

You need to get the 1st and 2nd derivatives of your equation. The 1st derivative gi ves you the slope of your curve at any point. Set your 1st derivative to zero and find what x is/are. At these x values the slope of your curve is zero, i.e. they are relative maxima or minima.

Next, get your 2nd derivative, plug-in the x-values derived previously and see if the 2nd derivative is negative or positive at these x-values. A negative value means it is concave down, a positive means it is concave up. You only need to focus on the values that make it concave down.

Next you need to get your points of inflection, x-values that make your 2nd derivative zero. This is where the slope of your curve changes from positive to negative. Your curve changes slope at three points (at the x-values found from the 1st derivative and from the x-values from the second derivative).

The valid x-values for your problem is from the point where the 2nd derivative is zero to negative infinity. Below is the mathematical solution:

f(x) = (x+5)(x-2)^2
= x^3 + x^2 - 16x + 20

dy/dx = 3x^2 +2x -16
set dy/dx = 0
then x = 2 or x = -8/3

d2y/dx2 = 6x + 2
at x = 2, d2y/dx2 = 14 (concave up)
at x = -8/3, d2y/dx2 = -14 (concave down)

set d2y/dx2 = 0
then x = -1/3 (Point of Inflection)

Valid x-values are: x = -1/3 to negative infinity

Answer 2

f(x) = (x+5)( (x^2) -4x +4) = (x^3) -4(x^2) +4x +5(x^2) -20x +20 = (x^3) +(x^2) -16x +20 f'(x) = 3(x^2) +2x -16 equalise this to 0 to acquire the table certain factors 3(x^2) +2x -16 = 0 (3x +8) (x -2)= 0 x= -8/3 , +2 to now it extremely is the min. do a second differentiation f"(x) = 6x +2 if the cost of this function is efficient, even as substituting the x with the values received, then this provides the x-coordinate of the min. so substituting x=2, f"(x) = 14 so the min. has an x-coordinate of two, even as the max. has an x-coordinate of -8/3. The graph is concave downward between those 2 values because the curve is shifting from a max. aspect to a min. aspect. so -8/3

Answer 3

you plug values in for y (f(x)) and it gives you the corresponding x's

Answer 4

why do people want some else to do their homework for them here?

Answer 5

conCAVE......Think about it.