Consider the function below. (Give your answers correct to three decimal places.

Question

Consider the function below. (Give your answers correct to three decimal places. If you need to use -infinity or infinity, enter -INFINITY or INFINITY.) h(x) = 7x^5 - 12x3^ + x (a) Find the intervals of increase.(Enter the intervals that contain smaller numbers first.) Find the intervals of decrease. (Enter the interval that contains smaller numbers first.) (b) Find the local minimum values. Find the local maximum values. (c) Find the inflection points. Find the intervals the function is concave up. (Enter the interval that contains smaller numbers first.) Find the intervals the function is concave down. (Enter the interval that contains smaller numbers first.) (d) Use this information to sketch the graph of the function. (Do this on paper. Your instructor may ask you to turn in this graph.)

Explanation / Answer

h(x) = 7x^5 - 12x^3 + x

Deriving :

h'(x) = 35x^4 - 36x^2 + 1 = 0 ---> to find the critical points

(35x^2 - 1)(x^2 - 1) = 0

x^2 = 1 and x^2 = 1/35

So, -1 , -1/sqrt35 , 1/sqrt35 , 1 are the critical values

So, this splits the number line into (-inf , -1) , (-1 , -1/sqrt35) , (-1/sqrt35 , 1/sqrt35) ,(1/sqrt35 , 1) and (1 , infinity)

Region 1 : (-inf , -1)
Testvalue = -2
h'(x) = 35x^4 - 36x^2 + 1
h'(-2) = 35(-2)^4 - 36(-2)^2 + 1
h(-2) = 417 --> posiitve --> increasing

Region2 : (-1 , -1/sqrt35) --> decreasing

Region 3 : (-1/sqrt35 , 1/sqrt35) --> increasing

Region 4 : (1/sqrt35 , 1) --> decreasing

Region 5 : (1 , infinity) --> increasing

So, increase : (-inf , -1) U (-1/sqrt35 , 1/sqrt35) U (1 , infinity) ----> ANSWER
Decrease : (-1 , -1/sqrt35) U (1/sqrt35 , 1) -----> ANSWER

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b) Lets plug in the critical numbers into the function
h(x) = 7x^5 - 12x^3 + x

h(-1) = 4
h(-1/sqrt(35)) = -0.112
h(1/sqrt(35)) = 0.112
h(1) = -4

So, local minimum values :
-1/sqrt(35) ---> ANSWER
1 --> ANSWER

Local maximum values :
-1 ---> ANSWER
1/sqrt(35) ---> ANSWER

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c) Inflection pts :
h'(x) = 35x^4 - 36x^2 + 1

Deriving again :

h''(x) = 140x^3 - 72x = 0

4x(35x^2 - 18) = 0

4x = 0 , 35x^2 - 18 = 0

x = 0 , x = +/- sqrt(18/35)

So, inflection points are :

-3sqrt(2/35) ---> ANSWER
0 ---> ANSWER
3sqrt(2/35) --> ANSWER

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d) Concavity :

From the inflections, the number line is split as (-infinity , -3sqrt(2/35)) , (-3sqrt(2/35)) , 0) , (0 , 3srt(2/35)) and (3sqrt(2/35)) , infinity)

Region 1 : (-infinity , -3sqrt(2/35))
Testvalue = -4
h''(x) = 140x^3 - 72x
h''(-4) = -8672 --> negative
So, concave down

So, by testing similarly, we get the concavity as follows :

Concave up :
(-3sqrt(2/35)) , 0) U (3sqrt(2/35)) , infinity) ---> ANSWER

Concave down :
(-infinity , -3sqrt(2/35)) U (0 , 3srt(2/35)) ---> ANSWER

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