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Sagot :
Answer:
[tex]\textsf{C is } \dfrac{-2 -\sqrt{2}}{8}\approx -0.4268[/tex]
[tex]\textsf{D is } \dfrac{-2 +\sqrt{2}}{8}\approx-0.0732[/tex]
(-∞, C]: concave up
[C, D]: concave down
[D, ∞): concave up
Step-by-step explanation:
Given function:
[tex]f(x)=x^2e^{8x}[/tex]
An inflection point is a point on the graph of the function where the concavity changes:
- A function f(x) is concave up on an interval if its second derivative f''(x) > 0 on that interval.
- A function f(x) is concave down on an interval if its second derivative f''(x) < 0 on that interval.
To find the inflection points of the given function, first find the second derivative f''(x).
Differentiate the function using the product rule.
[tex]\boxed{\begin{array}{c}\underline{\textsf{Product Rule fo Differentiation}}\\\\\textsf{If}\;y=uv\;\textsf{then:}\\\\\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}\end{array}}[/tex]
[tex]\textsf{Let }u = x^2 \implies \dfrac{\text{d}u}{\text{d}x}=2x[/tex]
[tex]\textsf{Let }v = e^{8x} \implies \dfrac{\text{d}v}{\text{d}x}=8e^{8x}[/tex]
Therefore:
[tex]f'(x)=x^2 \cdot 8e^{8x}+e^{8x}\cdot 2x \\\\\\ f'(x)=8x^2e^{8x}+2xe^{8x} \\\\\\ f'(x)=e^{8x}\left(8x^2+2x\right)[/tex]
Find the second derivative by differentiating again using the product rule.
[tex]\textsf{Let }u = e^{8x} \implies \dfrac{\text{d}u}{\text{d}x}=8e^{8x}[/tex]
[tex]\textsf{Let }v = 8x^2+2x \implies \dfrac{\text{d}v}{\text{d}x}=16x+2[/tex]
Therefore:
[tex]f''(x)=e^{8x} \cdot (16x+2)+(8x^2+2x) \cdot 8e^{8x} \\\\\\ f''(x)=16xe^{8x} +2e^{8x} + 64x^2e^{8x}+16xe^{8x} \\\\\\ f''(x)=64x^2e^{8x}+32xe^{8x} +2e^{8x} \\\\\\ f''(x)=e^{8x}\left(64x^2+32x+2\right)[/tex]
Now, set the second derivative equal to zero to find the inflection points.
[tex]f''(x)=0\\\\e^{8x}\left(64x^2+32x+2\right)=0[/tex]
Since [tex]e^{8x}\neq 0[/tex], we can solve:
[tex]64x^2+32x+2=0 \\\\2(32x^2+16x+1)=0 \\\\32x^2+16x+1=0[/tex]
Solve for x using the quadratic formula:
[tex]x=\dfrac{-16 \pm \sqrt{16^2-4(32)(1)}}{2(32)} \\\\\\ x=\dfrac{-16 \pm \sqrt{256-128}}{64} \\\\\\ x=\dfrac{-16 \pm \sqrt{128}}{64} \\\\\\ x=\dfrac{-16 \pm 8\sqrt{2}}{64} \\\\\\ x=\dfrac{-2 \pm \sqrt{2}}{8}[/tex]
Therefore, the two inflection points x = C and x = D where C ≤ D are:
[tex]\textsf{C is } \dfrac{-2 -\sqrt{2}}{8}\approx -0.4268[/tex]
[tex]\textsf{D is } \dfrac{-2 +\sqrt{2}}{8}\approx-0.0732[/tex]
[tex]\dotfill[/tex]
Now, determine the concavity on the given intervals by substituting a test value for x within each interval into the second derivative of the function.
For the interval (-∞, C], substitute the test point x = -1 into f''(x):
[tex]f''(-1)=e^{8(-1)}\left(64(-1)^2+32(-1)+2\right) \\\\ f''(-1)=e^{-8}\left(64-32+2\right) \\\\ f''(-1)=34e^{-8} \\\\ f''(-1)=0.0114057293... \\\\f''(-1) > 0[/tex]
As f''(-1) > 0, the function f(x) is concave up on the interval (-∞, C].
For the interval [C, D], substitute the test point x = -0.3 into f''(x):
[tex]f''(-0.3)=e^{8(-0.3)}\left(64(-0.3)^2+32(-0.3)+2\right) \\\\ f''(-0.3)=e^{-2.4}\left(5.76-9.6+2\right) \\\\ f''(-0.3)=-1.84e^{-2.4} \\\\ f''(-0.3)=-0.166921034... \\\\f''(-0.3) < 0[/tex]
As f''(-0.3) < 0, the function f(x) is concave down on the interval [C, D].
For the interval [D, ∞), substitute the test point x = 0 into f''(x):
[tex]f''(0)=e^{8(0)}\left(64(0)^2+32(0)+2\right) \\\\ f''(0)=e^{0}\left(0-0+2\right) \\\\ f''(0)=2 \\\\f''(0) > 0[/tex]
As f''(0) > 0, the function f(x) is concave up on the interval [D, ∞).
Therefore:
- (-∞, C]: concave up
- [C, D]: concave down
- [D, ∞): concave up
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