Question

section 2.7: second derivative and concavity
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question 5
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consider the function (f(x)=x^{2}e^{16x}).
(f(x)) has two inflection points at (x = c) and (x = d) with (c<d)
where (c) is
and (d) is
finally for each of the following intervals, tell whether (f(x)) is concave
((-infty,c)): concave up
((c,d)): concave down
((d,infty)) concave up
question help: video

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:

\(C=\frac{-2-\sqrt{2}}{16}\)
\(D=\frac{-2+\sqrt{2}}{16}\)