How do you determine whether the function #f(x)=x^3+3x^2+5x+7# is concave up or concave down and its intervals?

Question

Isabella Ackerman · Accepted Answer

Investigate the sign of the second derivative, #f''(x)# For #f(x)=x^3+3x^2+5x+7#, we have #f'(x)=3x^2+6x+5# and #f''(x) = 6x+6# #f''(x) = 0# at #x= -1#, so we check the sign of #f''# on each side of #-1# #f''(x)# is negative (that is #f''(x) < 0#) for #x < -1#. So the graph of #f# is concave down on #(-oo, -1)#. #f''(x) >0# for #x > -1#, so the graph of #f# is concave up on #(-1,oo)# The point #(-1, 4)# is the inflection point for the graph.

Charles Arocha · Answer

undefined To determine whether the function $ f(x) = x^3 + 3x^2 + 5x + 7 $ is concave up or concave down and its intervals, you can find the second derivative of the function and then analyze its sign.

The second derivative, denoted as $ f''(x) $, represents the concavity of the function. If $ f''(x) > 0 $ for a given interval, the function is concave up on that interval. If $ f''(x) < 0 $ for a given interval, the function is concave down on that interval.

To find $ f''(x) $, you first find the first derivative $ f'(x) $, then take the derivative of that result to find $ f''(x) $.

Given $ f(x) = x^3 + 3x^2 + 5x + 7 $, the first derivative $ f'(x) $ is $ f'(x) = 3x^2 + 6x + 5 $.

Taking the derivative of $ f'(x) $ yields the second derivative $ f''(x) = 6x + 6 $.

To determine the intervals of concavity, you need to find the critical points of $ f''(x) $ by setting $ f''(x) = 0 $ and solving for $ x $.

$ 6x + 6 = 0 $

$ x = -1 $

Now, test the intervals defined by the critical point $ x = -1 $ and any other relevant points in the second derivative.

For $ x < -1 $, choose a test point such as $ x = -2 $:

$ f''(-2) = 6(-2) + 6 = -12 + 6 = -6 $ (negative, concave down)

For $ -1 < x < \infty $, choose a test point such as $ x = 0 $:

$ f''(0) = 6(0) + 6 = 6 $ (positive, concave up)

Therefore, the function $ f(x) = x^3 + 3x^2 + 5x + 7 $ is concave down for $ x < -1 $ and concave up for $ -1 < x < \infty $.