(I) T or F: The function is concave up at x=1/1O? True False (m) T or F: The fun

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(I) T or F: The function is concave up at x=1/1O? True False (m) T or F: The function is concave up at x=1? True False (n)The value of Iimx rightarrow infinity f'(x) is .(If you need to use -infinity or infinity .) (o) The graph of y=f(x) is best represented by which of the following pictures (select A, B, C or D): A B C D Question For this problem, consider the function Y =f(x)= x^4.x on the domain of all positive numbers; i.e. (O,infinity) . (If you need to use -infinity or infinity, enter -infinity or infinity.) (b) T or F: The function has no x intercepts True False (c) T or F: The y axis is a vertical asymptote True False (d) The value of Iimitx rightarrow infinityx^4.x is .(If you need to use -infinity or infinity, enter -infinity or infinity.) (e) What is the formula for f?(x)= (f) How many critical numbers are there for f(x)? (h) The smallest critical number is (I) The largest interval in the domain on which the function is increasing is ( , ). (If you need to use -infinity or infinity, enter -infinity or infinity.) (j) There is a local minimum occurring at ( , ) on the graph. (k) The formula for f^'(x)=

Explanation / Answer

y = x^(4x)

ln(y) = 4x*ln(x)

ln(y) = 4ln(x) / (1/x)

When x ---> 0+, 4ln(0+) --> -infinity
And 1/0+ ---> +infinity

So, infinity by infinity form and thus, we can use L Hospital's Rule

ln(y) = 4ln(x) / (1/x)

Deriving :

ln(y) = (4/x) / (-1/x^2)

ln(y) = 4/x * -x^2

ln(y) = -4x

ln(y) = -4*0 = 0

ln(y) = 0

y = e^0

y = 1 ---> ANSWER

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b) True

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c) True

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d)

lim x ---> infinity :

ln(y) = 4x*ln(x)

ln(y) = 4ln(x) / (1/x)

Deriving :

ln(y) = 4/x / (-1/x^2)

ln(y) = x/4

ln(y) = infinity/4

ln(y) = infinity

y = e^(infinity)

y = +infinity ---> ANSWER

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e)

f(x) = x^(4x)
y = x^(4x)
ln(y) = 4x*ln(x)

Deriving :

y'/y = 4 + 4ln(x)

y' = y(4 + 4ln(x))

y' = x^(4x) * (4 + 4ln(x)) ---> ANSWER

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f) Critical numbers :

x^(4x) * (4 + 4ln(x)) = 0

x^(4x) = 0 and 4 + 4ln(x) = 0

x^(4x) = 0 ---> no solution

4 + 4ln(x) = 0

ln(x = -1

x = e^-1

x = 1/e

So, only one critical number ----> ANSWER

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h) 1/e --> ANSWER

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i)

Increases over (1/e , infinity) ---> ANSWER

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j)

When x = 1/e, y = (1/e)^(4*1/e) --> y = (1/e)^(4/e) = 0.2295767771002993

Local minimum at (1/e , 0.2295767771002993) ---> ANSWER

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k) f''(x) :

f'(x) = x^(4x) * (4 + 4ln(x))

f''(x) = x^(4x) * (4/x + (4ln(x) + 4)^2) ---> ANSWER

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L)

x = 1/10

Lets plug in x = 1/10 into the f'' found above

f''(1/10) = 0.1^(4*0.1) * (4/0.1 + (4ln(0.1) + 4)^2) = 26.7319596736269926 --> positive

So, since it is positive, it is concave up

So, TRUE ---> ANSWER

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M)

x = 1 :

f''(1) = 1^(4*1) * (4/1 + (4ln(1) + 4)^2) = 20 --> positive

So, TRUE ---> ANSWER

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N)

f'(x) = x^(4x) * (4 + 4ln(x))

f'(x) = 4x^(4x) + 4x^(4x)*ln(x)

From the graph this is a monotonically increasing function

So, as lim x --> infinity, f'(x) will also tend to infinity

+infinity ----> ANSWER

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O)

Plot B ---> ANSWER

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