section 2.7: second derivative and concavity\nscore: 30.67/80 answered: 3/8\nquestion 5\nscore on last try…
section 2.7: second derivative and concavity\nscore: 30.67/80 answered: 3/8\nquestion 5\nscore on last try: 4 of 10 pts. see details for more.\nat least one scored part is incorrect. jump to first changeable incorr\n> next question get a similar question you can retry th\nconsider the function (f(x)=x^{2}e^{16x}).\n(f(x)) has two inflection points at (x = c) and (x = d) with (c<d)\nwhere (c) is\nand (d) is\nfinally for each of the following intervals, tell whether (f(x)) is concave\n((-infty,c)): concave up\n((c,d)): concave down\n((d,infty)) concave up\nquestion help: video
Answer
Explanation:
Step1: Find the first - derivative using product rule
The product rule states that if (y = uv), where (u = x^{2}) and (v=e^{16x}), then (y'=u'v + uv'). (u' = 2x) and (v'=16e^{16x}), so (f'(x)=2x e^{16x}+16x^{2}e^{16x}=2xe^{16x}(1 + 8x)).
Step2: Find the second - derivative using product rule again
Let (u = 2x(1 + 8x)=2x+16x^{2}) and (v = e^{16x}). (u'=2 + 32x) and (v'=16e^{16x}). (f''(x)=(2 + 32x)e^{16x}+16(2x + 16x^{2})e^{16x}=e^{16x}(2 + 32x+32x + 256x^{2})=e^{16x}(256x^{2}+64x + 2)).
Step3: Set the second - derivative equal to zero
Since (e^{16x}\gt0) for all real (x), we solve (256x^{2}+64x + 2 = 0). Divide through by 2: (128x^{2}+32x + 1 = 0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) with (a = 128), (b = 32), and (c = 1). (x=\frac{-32\pm\sqrt{32^{2}-4\times128\times1}}{2\times128}=\frac{-32\pm\sqrt{1024 - 512}}{256}=\frac{-32\pm\sqrt{512}}{256}=\frac{-32\pm16\sqrt{2}}{256}=\frac{-2\pm\sqrt{2}}{16}). So (C=\frac{-2-\sqrt{2}}{16}) and (D=\frac{-2 + \sqrt{2}}{16}).
Answer:
(C=\frac{-2-\sqrt{2}}{16}) (D=\frac{-2+\sqrt{2}}{16})