(1) (a) Find the critical points for f(x)=x^2- 7x + 17. (b) Determine the interv

Question

(1) (a) Find the critical points for f(x)=x^2- 7x + 17. (b) Determine the intervals where f is increasing or decreasing. (c) Classify each critical point as local maximum, local minimum, or neither one. (2) (a) Find the critical points for f(x) = x^2 ? x^4. (b) Determine the intervals where f is increasing or decreasing. (c) Classify each critical point as local maximum, local minimum, or neither one. (d) Determine the intervals where f is concave up and where it is concave down. (e) Determine points of inflection for f. (3) (a) Find the critical points for f(x) = x +cosx, 0 less than equal to x less than equal to 2pi. (b) Determine the intervals where f is increasing or decreasing. (c) Classify each critical point as local maximum, local minimum, or neither one. (d) Determine the intervals where f is concave up and where it is concave down. (e) Determine points of inflection for f. (4) (a) Determine the intervals where f(x)= x^12/5 is concave up and concave down. (b) Determine points of inflection for f. (5) (Bonus question): Let f(x)= sin^2 x +cos x. Find the critical numbers for f in the interval [0,pi], and classify each one as a local max, local min, or neither one using the second derivative test.

Explanation / Answer

1)

a) f(x) = x^2 - 7x + 17
f'(x) = 2x - 7 = 0 --> x = 7/2
When x = 7/2, y = (7/2)^2 - 7(7/2) + 17 = 19/4

(7/2 , 19/4)

b) Increasing over (7/2 , infinity)
Decreasing over (-infinity , 7/2)

c)

From increasing-decreasing info, since "decreasing changes to increasing" at x = 7/2,
the point (7/2 , 19/4) is a MINIMUM

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2)

a) f(x) = x^2 - x^4
f'(x) = 2x - 4x^3 = 0
2x(1 - 2x^2) = 0
x = 0 , 1/sqrt2 and -1/sqrt2

b)
Increasing : (-inf , -1/sqrt2) U (0 , 1/sqrt2)
Decreasing : (-1/sqrt2 , 0) U (1/sqrt2 , infinity)

c)
x = -1/sqrt2 and x = 1/sqrt2 are maxima
x = 0 is a minima

e)
f''(x) = 2 - 12x^2 = 0
x = 1/sqrt6 and -1/sqrt6

d)
Using inflection points 1/sqrt6 and -1/sqrt6, we can find the intervals of concavity

Regions of consideration are (-inf , -1/sqrt6) , (-1/sqrt6 , 1/sqrt6) and (1/sqrt6 , inf)

Region 1 : (-inf , -1/sqrt6)
Testvalue 1 : -2

f''(-2) = 2 - 12(-2)^2 ---> negative --> concave down

Region 2 : (-1/sqrt6 , 1/sqrt6)
Testvalue 2 : 0

f''(0) = 2 - 12(0)^2 = 2 ---> positive --> concave up

Region 3 : concave down

So, concave down over (-inf , -1/sqrt6) U (1/sqrt6 , inf)
And concave up over (-1/sqrt6 , 1/sqrt6)

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3)

f(x) = x + cosx

a) f'(x) = 1 - sinx = 0 --> sinx = 1 ---> x = pi/2

b) Increasing-decreasing :

Critical number is pi/2

So, (0 , pi/2) and (pi/2 , 2pi) are the regions of interest

Region 1 : (0 , pi/2)
Testvalue 1 = pi/4
f'(pi/4) = 1 - sin(pi/4) --> positive --> increasing

Region 2 : (pi/2 , 2pi)
testvalue 2 = pi
f'(pi) = 1 - sin(pi) = 1 --> positive --> increasing

So, increases over [0 , 2pi]
Decreases nowhere

c) x = pi/2 is neither a max nor a min

e)

f'(x) = 1 - sinx
f''(x) = -cosx = 0 --> x = pi/2 and 3pi/2

d)

Concavity :
(0 , pi/2) , (pi/2 , 3pi/2) and (3pi/2 , 2pi)

Region 1 : (0,pi/2)
testvalue 1 : pi/4

f''(pi/4) = -cos(pi/4) --> negative --> concave down

Similarly, region 2 will be concave up

Region 3 will be concave down

Concave up over (pi/2 , 3pi/2)
Concave down over [0 , pi/2) U (3pi/2 , 2pi]

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4)

f(x) = x^(12/5)
f'(x) = 12/5 * x^(7/5)
f''(x) = (12/5)(7/5) * x^(2/5) = 0

x^(2/5) = 0
x = 0

b) x = 0

a) Only inflection point is x = 0
So, this splits the line into (-inf , 0) and (0 , inf)

Region 1 : (-inf , 0)
Testvalue 1 : -1

f''(-1) = 84/25 * (-1)^(2/5) --> negative --> concave down

Region 2 : Concave up

So, concave down over (-inf , 0)
Concave up over (0 , inf)

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5)

f(x) = sin^2x + cosx ; [0 , pi]
f'(x) = 2sinxcosx - sinx = 0

sinx(2cosx -1) = 0

sinx = 0 and cosx = 1/2

x = 0 , pi and x = pi/6

f''(x) = -2sin^2x + 2cos^2x - cosx

f''(0) = 0 + 2 - 1 = 1 --> positive value --> minimum

f''(pi) = 0 + 2 - (-1) --> positive value ---> minimum

f''(pi/6) = -2(1/2)^2 + 2(sqrt3/2)^2 - sqrt3/2
= -1/2 + 3/2 - sqrt3/2
= 1 - sqrt3/2 --> positive value ---> minimum

So, all of the values x = 0 , pi/6 and pi are local minima

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