For what values of x is #f(x)= -9x^3 + 4 x^2 + 7x -2 # concave or convex?

A function is concave (or concave down) where its derivative is decreasing. Graphically, a concave region looks like a cave (or cut from a cave shape). (Quick example: the function #f(x)="–"x^2# is concave everywhere, since it never curves upward.)

A function is convex (or concave up) where its derivative is increasing. Graphically, a convex region looks like a V (or cut from a V shape). (The function #f(x)=x^2# is convex everywhere.)

To find these regions, we need to analyze the behaviour of the function's derivative. Hence, we need to find the derivative of the derivative: #f''(x)#.

Given #f(x)="–"9x^3+4x^2+7x-2#, we have

#f'(x)="–"27x^2+8x+7#

#f''(x)="–"54x+8#.

Just like how we know #f(x)# is increasing when #f'(x)>0#, we know #f'(x)# is increasing when #f''(x)>0#. Using this, we find:

#f''(x)="–"54x+8 > 0#

#=>"                 –"54x>"–8"#

#=>"                       "x<8/54=4/27#.

So #f(x)# is convex (concave up) for #x in ("–"oo," "4/27)#. This means #x=4/27# is an inflection point—a point at which the direction of concavity changes.

Finding where #f(x)# is concave (concave down) means finding where #f''(x)<0#, but that's equivalent to swapping #># signs for#<# signs from when we found #f''(x) > 0#. So our answer will be #f(x)# is concave for #x in (4/27," "oo)#.

As we come from #"–"oo#, the slope (of the tangent line) of the graph is increasing, right up until the inflection point, where the slope begins to decrease.

#f(x)=-9x^3+4x^2+7x-2#

#f'(x)=-27x^2+8x+7#

#-27x^2+8x+7=0#

#Delta=8^2-4*(-27)*(7)=820#

As, #Delta>0#, there are 2 real roots

#x_1=(-8+sqrt820)/(2*-27)=-0.382#

#x_2=(-8-sqrt820)/(2*-27)=0.678#

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaa)##-0.382##color(white)(aaaa)##0.678##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-x_1##color(white)(aaaaaa)##+##color(white)(aaaaaaa)##-##color(white)(aaaaaa)##-#

#color(white)(aaaa)##x-x_2##color(white)(aaaaaa)##-##color(white)(aaaaaaa)##-##color(white)(aaaaaa)##+#

#color(white)(aaaa)##f'(x)##color(white)(aaaaaaa)##-##color(white)(aaaaaaa)##+##color(white)(aaaaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##↘##color(white)(aaaaaaa)##↗##color(white)(aaaaaa)##↘#

#f(x)# is convex when #x in ]-oo,0.678]#

and concave when #x in [-0.382,+oo[#

To determine the concavity of the function \( f(x) = -9x^3 + 4x^2 + 7x - 2 \), you need to find its second derivative and then analyze its sign. The second derivative of \( f(x) \) is denoted as \( f''(x) \) and can be found by taking the derivative of the first derivative of \( f(x) \). So, first, find the first derivative, \( f'(x) \), and then differentiate it again to find \( f''(x) \). Given \( f(x) = -9x^3 + 4x^2 + 7x - 2 \): 1. Find \( f'(x) \): \[ f'(x) = \frac{d}{dx}(-9x^3 + 4x^2 + 7x - 2) \] \[ f'(x) = -27x^2 + 8x + 7 \] 2. Find \( f''(x) \): \[ f''(x) = \frac{d}{dx}(-27x^2 + 8x + 7) \] \[ f''(x) = -54x + 8 \] Now, to determine concavity or convexity: - If \( f''(x) > 0 \), the function is concave up (convex). - If \( f''(x) < 0 \), the function is concave down. Setting \( f''(x) \) to zero gives you the inflection point(s) where the concavity might change. \[ -54x + 8 = 0 \] \[ x = \frac{8}{54} = \frac{4}{27} \] So, the function might change concavity at \( x = \frac{4}{27} \). Now, evaluate \( f''(x) \) for \( x < \frac{4}{27} \) and \( x > \frac{4}{27} \) to determine the concavity in these intervals. - For \( x < \frac{4}{27} \), \( f''(x) < 0 \), hence the function is concave down. - For \( x > \frac{4}{27} \), \( f''(x) > 0 \), hence the function is concave up (convex). Therefore, the function \( f(x) = -9x^3 + 4x^2 + 7x - 2 \) is concave down for \( x < \frac{4}{27} \) and concave up (convex) for \( x > \frac{4}{27} \).