What are the intervals which f increases and decreases, where f is concave up and down, and coordinates of inflection points? #x^4-5x^3+9x^2#

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VNVDVI
Apr 7, 2018

Decreasing: #(-oo, 0)# Increasing: #(0, oo)# Minimum: #(0,0)# Concave up: #(-oo, 1), (3/2, oo)# Concave down: #(1, 3/2)# Inflection point: #(3/2,189/16)#

Explanation:

Take the first derivative, set equal to zero, and solve for #x# to obtain critical values. We would also have to see where the first derivative doesn't exist; however, this is a polynomial and will therefore have a continuous derivative.

#f'(x)=4x^3-15x^2+18x#

#4x^3-15x^2+18x=0#

#x(4x^2-15x+18)=0#

That quadratic cannot be factored further; thus, #x=0# is the only critical value.

We must then determine whether #f'(x)# is positive or negative in the intervals #(-oo,0), (0, oo)#.

#(-oo,0): f'(-1)=4(-1)^3+15(-1)^2-18=4+15-18<0#

A negative first derivative means a decreasing function, so, #f(x)# decreases on #(-oo, 0)#

#(0, oo): f'(1)=4-15+18>0#

A positive first derivative means an increasing function, so, #f(x)# increases on #(0, oo)#.

We then have a local minimum at #(0, f(0))# as we change from decreasing to increasing at that point.

#f(0)=0^4-5(0^3)+9(0^2)=0#

#(0,0)# is the minimum.

Take the second derivative, set equal to zero, and solve. Again, it will be continuous, so no need to check for points where it does not exist.

#f''(x)=12x^2-30x+18#

#12x^2-30x+18=0#

#6(x-1)(2x-3)=0#

#x=1, x=3/2#

We must determine whether the second derivative is positive or negative on the intervals #(-oo, 1), (1, 3/2), (3/2, oo)#

#(-oo,1):f''(0)=12(0^2)-30(0)+18=18>0#

A positive second derivative entails upward concavity.

#f(x)# is concave up on #(-oo, 1).#

#(1, 3/2): f''(5/4)=12(5/4)^2-30(5/4)+18=300/16-150/4+18=300/16-600/16+288/16<0#

A negative second derivative entails downward concavity.

#f(x)# is concave down on #(1, 3/2).#

#(3/2, oo): f''(2)=12(2^2)-30(2)+18=48-60+18>0#

Concave up on #(3/2, oo)#

There will be an inflection point at #(3/2, f(3/2))# due to changing concavity at that point.

#f(3/2)=(3/2)^4-5(3/2)^3+9(3/2)^2=81/16-135/4+81/2=81/16-540/16+648/16=189/16#

The inflection point is at #(3/2, 189/16)#

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