Question

Find the relative maximum / minimum / points of inflection of the following functions:

y=x3+6x2+7

y=2x(1−2x)

y=ln(2x−5)−lnx

y= 13 x3 −15x2 +300x+150

Answer 1

1) find the derivative:

$y'=(x3+6x2+7)'=3x^2+12x$

3x(x+4)=x

x₁ = 0

x₂ = -4

In the vicinity of the point x = -4, the derivative of the function changes sign from (+) to (-). Therefore, the point x = -4 is the maximum point. In the vicinity of the point x = 0, the derivative of the function changes sign from (-) to (+). Therefore, the point x = 0 is the minimum point.

the second derivative:

$(3x^2+12x)''=6x+12$

6x+12=0

x=-2 point of inflection

2)find the derivative:

$y'=2x(1−2x)=(2x-4x^2)=2-8x$

2-8x=0

x=0.25

In the vicinity of the point x = 0.25 the derivative of the function changes sign from (+) to (-). Therefore, the point x = 0.25 is the maximum point.

There are no inflection points, since the second derivative is -8, and this is not equal to 0

3)find the derivative:

$y'=(Ln(2x-5)-Lnx)'=\frac{5}{2x^2-5x}$

$\frac{5}{2x^2-5x}=0$

To do this, we equate the derivative to zero

There are no roots for this equation.

the second derivative:

$(\frac{5}{2x^2-5x})''=\frac{25-20x}{x^2(2x-5)^2}$

$\frac{25-20x}{x^2(2x-5)^2}=0$

x=1.25 point of inflection

4)find the derivative:

$y'=(13x^3-15x^2+300x+150)'=39x^2- 30x+300$

$13x^2-10x+100=0$

There are no roots for this equation.

the second derivative:

$(39x^2-30x+300)'=78x-30$

78x-30=0

x=0.38 point of inflection