Question

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 and 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

Answer 1

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.

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