How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for #y=(2x+3)^2(x+1)^2# for #[-10,0]#?

Question

Scarlett Allison · Accepted Answer

Explanation below
graph{(2x+3)^2(x+1)^2 [-2.035, 0.098, -0.503, 0.5646]}

The domain: #[-10,0]# #f^'(x)=4(2x+3)(x+1)^2+2(2x+3)^2(x+1)# #f^'(x)=2(2x+3)(x+1)[2(x+1)+2x+3]=2(2x+3)(x+1)(4x+5)# #2x+3=0quad^^quadx+1=0quad^^quad4x+5=0# #x=-3/2quad^^quadx=-1quad^^quadx=-5/4# #x in [-10,-3/2] hArr f^'<0 => #f goes down #x=-3/2 =># minimum #f(-3/2)=0# #x in [-3/2,-5/4] hArr f^'>0 => #f goes up #x=-5/4 =># local maximum #x in [-5/4,-1] hArr f^'<0 => #f goes down #x=-1 =># minimum #f(-1)=0# #x in [-1,0] hArr f^'>0 => #f goes up Global maximum: #f(-10)=(-20+3)^2(-10+1)^2=23409# Cocavity: #f^'(x)=4(2x+3)(x^2+2x+1)+2(4x^2+12x+9)(x+1)# #f^'(x)=2(8x^3+30x^2+37x+15)# #f^''(x)=2(24x^2+60x+37)# #x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(-60+-sqrt(60^2-4*24*37))/(2*24)# #x_(1,2)=(-15+-sqrt(3))/(12)# #x_(1)=-1.394...# #x_(2)=-1.105...# #x in [-10,x_1) hArr f^''>0 =># Convex #x in (x_1,x_2) hArr f^''<0 =># Concave #x in (x_2,0] hArr f^''>0 =># Convex

Christopher Allers · Answer

undefined To find the x-coordinates of all points of inflection, we need to find the second derivative of the function $ y = (2x+3)^2(x+1)^2 $ and then solve for where the second derivative equals zero or is undefined.

The first derivative is:

$$ y' = 2(2x+3)(x+1)^2 + 2(x+1)(2x+3)^2 $$

The second derivative is:

$$ y'' = 2(2)(x+1)^2 + 2(2x+3) \cdot 2(x+1) + 2(2x+3) \cdot 2(x+1) + 2(x+1)(2)(2x+3) $$

$$ y'' = 4(x+1)^2 + 4(2x+3) + 4(2x+3) + 4(x+1)(2) $$

$$ y'' = 4(x^2 + 2x + 1) + 8x + 12 + 8x + 12 + 8x + 12 $$

$$ y'' = 4x^2 + 8x + 4 + 8x + 12 + 8x + 12 $$

$$ y'' = 4x^2 + 24x + 28 $$

Now, we need to solve $ y'' = 0 $ to find the points of inflection:

$$ 4x^2 + 24x + 28 = 0 $$

This quadratic equation doesn't factor easily, so we can use the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Plugging in the values $ a = 4 $, $ b = 24 $, and $ c = 28 $, we get:

$$ x = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 4 \cdot 28}}{2 \cdot 4} $$

$$ x = \frac{-24 \pm \sqrt{576 - 448}}{8} $$

$$ x = \frac{-24 \pm \sqrt{128}}{8} $$

$$ x = \frac{-24 \pm 8\sqrt{2}}{8} $$

$$ x = -3 \pm \sqrt{2} $$

So, the x-coordinates of the points of inflection are $ -3 + \sqrt{2} $ and $ -3 - \sqrt{2} $.

To find the discontinuities, we need to check where the function is undefined. Since the function $ y = (2x+3)^2(x+1)^2 $ is polynomial, it's defined for all real numbers. Therefore, there are no discontinuities in this function.

Finally, to determine the open intervals of concavity, we need to analyze the sign of the second derivative. Since the second derivative is a positive quadratic, it's positive for all real values of $ x $, meaning the function is concave up for all values in the interval $[-10, 0]$.