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mario62 [17]
3 years ago
8

Question 7 (10 points) Given f(x)=(x−2)2(x−4)2 , determine a. interval where f(x) is increasing or decreasing, b. local minima a

nd maxima of f(x) c. intervals where f(x) is concave up and concave down, and d. the inflection points of f(x) . Sketch the curve, and then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculato
Mathematics
1 answer:
Bad White [126]3 years ago
4 0

Answer:

b. x=2,3,4

c.

x=2 is a local minimum

x=3 is a local maximum

x=4 is a local minimum

d.

x_1=3+\frac{\sqrt{12}}{6}\\x_2=3-\frac{\sqrt{12}}{6}

Step-by-step explanation:

a. You have the function

f(x)=(x-2)^2(x-4)^2

to find the intervals you have to take into account the roots of f(x), in this case you have the roots

x_{1,2}=2\\x_{3,4}=4

- for x < 2, for example x=1:

f(1)=(1-2)^2(1-4)^2>0

f(x) decreases from infinity to zero

- for 2 < x < 4, for example x=3:

f(3)=(3-2)^2(3-4)^2>0

between x=2 and x=3 f(x) increases. But between x=3 and x=4 f(x) decreases because f(4)=0.

- for x > 4:

f(5)>0

f(x) increases for x > 4.

b. You have to compute the derivative to find local minima and maxima

f'(x)=2(x-2)(x-4)^2+2(x-2)^2(x-4)=2(x-2)(x-4)[(x-4)+(x-2)]\\f'(x)=2(x-2)(x-4)(2x-6)

and by taking f'(x)=0:

f'(x)=0\\f'(x)=(x-2)(x-4)(x-3)=0\\\\x_1=2\\x_2=4\\x_3=3

Local minima and maxima are found by evaluating the roots of f'(x) in the second derivative f''(x)

f''(x)=2[(x-4)(2x-6)+(x-2)(2x-6)+(x-2)(x-4)(2)]\\f''(x)=2[2x^2-14x+24+2x^2-10x+12+2x^2-12x+16]\\f''(x)=12x^2-72x+104

Hence

f''(2)=12(2)^2-72(2)+104=8>0\\f''(3)=12(3)^2-72(3)+104=-40

x=2 is a local minimum

x=3 is a local maximum

x=4 is a local minimum

c.

to find the concavity you have to find the inflection points

f''(x)=0\\f''(x)=12x^2-72x+104=0\\3x^2-18x+26=0\\x=\frac{18+-\sqrt{324-4(3)(26)}}{6}\\x_1=3 + \frac{\sqrt{12}}{6}\\x_2=3 - \frac{\sqrt{12}}{6}\\

Hence:

for -infinity < x < x1 --> concave up

for x1 < x < x2 --> concave down

for x2 < x < infinity --> concave up

d.

x_1=3+\frac{\sqrt{12}}{6}\\x_2=3-\frac{\sqrt{12}}{6}

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