Can someone work this out please??? Thank you!
(a) Find the interval on which f is increasing.
(b) Find the interval on which f is decreasing.
(c) Find the local minimum value of f.
(d) Find the inflection point.
(e) Find the interval on which f is concave up.
(f) Find the interval on which f is concave down.
(a) Find the interval on which f is increasing.
(b) Find the interval on which f is decreasing.
(c) Find the local minimum value of f.
(d) Find the inflection point.
(e) Find the interval on which f is concave up.
(f) Find the interval on which f is concave down.
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to find the if the function increasing or decreasing, we need to find the first derivative and equal it to zero as the following:
f(x) = x^4 In(x)
f ' (x) = x^4 * (1/x) + ln(x) * 4x^3 ===>equal it to zero
0 = x^3 + ln(x) * 4x^3
x^3 * ( 1 + 4ln(x) ) = 0
x^3 = 0 ====> x = 0 (critical point )
1 + 4ln(x) = 0 ====> 4ln(x) = -1 ====> ln(x) = -1/4 ===> x = e^(-1/4) (critical point )
to find where the function is increasing or decreasing we need to check before and after these numbers in the first derived equation as the following:
f ' (x) = x^3 * ( 1 + 4ln(x) ) ====> no zero and negative numbers because of the divergence of ln
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____________________________
0.............e^(-1/4)
increasing ( e^(-1/4) , ∞ )
decreasing ( 0 , e^(-1/4) )
to find maximum or minimum, we need to plug the critical numbers in the original equation:
f(0) = 0^4 In(0) = ====> no go
f( e^(-1/4) ) = (e^(-1/4))^4 In(e^(-1/4)) ≈ -0.09 ( minimum ) ( critical value )
to find where the concavity, concaving up or down, we need to find the second derivative and equal it to zero:
f ''(x) = 3x^2 + ln(x) * 12x^2 + 4x^3 * (1/x) ===>equal it to zero
0 = 3x^2 + ln(x) * 12x^2 + 4x^2
0 = 7x^2 + ln(x) * 12x^2
0 = x^2 ( 7 + 12 * ln(x) )
x^2 = 0 =====> x = 0 ( points of inflection )
7 + 12 ln(x) = 0 ====> 12ln(x) = -7 ====> ln(x) = -7/12 ====> x = e^(-7/12) ( point of inflection )
to find the concavities, we need to check before and after these points of inflection into the second derived equation as the following:
f '' (x) = x^2 ( 7 + 12 * ln(x) ) ====> no zero and negative numbers because of the divergence of ln
---------------++++++++
____________________________
0.............e^(-7/12)
concave up ( e^(-7/12) , ∞ )
concave down ( 0 , e^(-7/12) )
to find the inflection point, we need to plug the points of inflection into the original equation as the following:
f(0) = 0^4 In(0 = =====> no
f(e^(-7/12)) = ( e^(-7/12) )^4 In( e^(-7/12) ) ≈ -0.07 ( inflection point )
f(x) = x^4 In(x)
f ' (x) = x^4 * (1/x) + ln(x) * 4x^3 ===>equal it to zero
0 = x^3 + ln(x) * 4x^3
x^3 * ( 1 + 4ln(x) ) = 0
x^3 = 0 ====> x = 0 (critical point )
1 + 4ln(x) = 0 ====> 4ln(x) = -1 ====> ln(x) = -1/4 ===> x = e^(-1/4) (critical point )
to find where the function is increasing or decreasing we need to check before and after these numbers in the first derived equation as the following:
f ' (x) = x^3 * ( 1 + 4ln(x) ) ====> no zero and negative numbers because of the divergence of ln
---------------++++++++
____________________________
0.............e^(-1/4)
increasing ( e^(-1/4) , ∞ )
decreasing ( 0 , e^(-1/4) )
to find maximum or minimum, we need to plug the critical numbers in the original equation:
f(0) = 0^4 In(0) = ====> no go
f( e^(-1/4) ) = (e^(-1/4))^4 In(e^(-1/4)) ≈ -0.09 ( minimum ) ( critical value )
to find where the concavity, concaving up or down, we need to find the second derivative and equal it to zero:
f ''(x) = 3x^2 + ln(x) * 12x^2 + 4x^3 * (1/x) ===>equal it to zero
0 = 3x^2 + ln(x) * 12x^2 + 4x^2
0 = 7x^2 + ln(x) * 12x^2
0 = x^2 ( 7 + 12 * ln(x) )
x^2 = 0 =====> x = 0 ( points of inflection )
7 + 12 ln(x) = 0 ====> 12ln(x) = -7 ====> ln(x) = -7/12 ====> x = e^(-7/12) ( point of inflection )
to find the concavities, we need to check before and after these points of inflection into the second derived equation as the following:
f '' (x) = x^2 ( 7 + 12 * ln(x) ) ====> no zero and negative numbers because of the divergence of ln
---------------++++++++
____________________________
0.............e^(-7/12)
concave up ( e^(-7/12) , ∞ )
concave down ( 0 , e^(-7/12) )
to find the inflection point, we need to plug the points of inflection into the original equation as the following:
f(0) = 0^4 In(0 = =====> no
f(e^(-7/12)) = ( e^(-7/12) )^4 In( e^(-7/12) ) ≈ -0.07 ( inflection point )