Determine the x-coordinates of the points of inflexion of the following function:
f\left(x\right)=5x^{3}-30x^{2}+25x-10
x=a is a point of inflexion for a function f if:
- f' is differentiable in x=a
- f''\left(a\right)=0
- f'' changes its sign in x=a
Calculate f'\left(x\right) :
f'\left(x\right)=\left(5x^{3}-30x^{2}+25x-10\right)'=15x^{2}-60x+25
Calculate f''\left(x\right) :
f''\left(x\right)=\left(15x^{2}-60x+25\right)'=30x-60
Solve the equation f''\left(x\right)=0 :
30x-60=0
x=2
The sign of f'' is the following:
The point of inflexion for the function f is x=2.
Determine the x-coordinates of the points of inflexion of the following function:
f\left(x\right)=x^{2}e^{3x}
x=a is a point of inflexion for a function f if:
- f' is differentiable in x=a
- f''\left(a\right)=0
- f'' changes its sign in x=a
Calculate f'\left(x\right) :
f'\left(x\right)=\left(x^{2}e^{3x}\right)'=\left(x^{2}\right)'e^{3x}+x^{2}\left(e^{3x}\right)'=2x\cdot e^{3x}+x^{2}\cdot e^{3x}\cdot3=\left(3x^{2}+2x\right)e^{3x}
Calculate f''\left(x\right) :
f''\left(x\right)=\left(\left(3x^{2}+2x\right)e^{3x}\right)'=\left(3x^{2}+2x\right)'e^{3x}+\left(3x^{2}+2x\right)\cdot\left(e^{3x}\right)'=\left(6x+2\right)e^{3x}+\left(3x^{2}+2x\right)\cdot e^{3x}\cdot3=\left(9x^{2}+12x+2\right)e^{3x}
Solve the equation f''\left(x\right)=0 :
\left(9x^{2}+12x+2\right)e^{3x}=0
e^{3x}=0 has no solution;
9x^{2}+12x+2=0 has solutions:
x_{1}=\dfrac{-2-\sqrt{2}}{3} and x_{2}=\dfrac{-2+\sqrt{2}}{3}
The sign of f'' is the following:
The points of inflexion for the function f are x=\dfrac{-2-\sqrt{2}}{3} and x=\dfrac{-2+\sqrt{2}}{3}.
Determine the x-coordinates of the points of inflexion of the following function:
f\left(x\right)=x^{3}-5x^{2}+7x-3
x=a is a point of inflexion for a function f if:
- f' is differentiable in x=a
- f''\left(a\right)=0
- f'' changes its sign in x=a
Calculate f'\left(x\right) :
f'\left(x\right)=\left(x^{3}-5x^{2}+7x-3\right)'=3x^{2}-10x+7
Calculate f''\left(x\right) :
f''\left(x\right)=\left(3x^{2}-10x+7\right)'=6x-10
Solve the equation f''\left(x\right)=0 :
6x-10=0
x=\dfrac{5}{3}
Since the sign of f'' changes in x=\dfrac{5}{3}, we conclude that x=\dfrac{5}{3} is an inflexion point.
The points of inflexion for the function f is x=\dfrac{5}{3}.
Determine the x-coordinates of the points of inflexion of the following function:
f\left(x\right)=\ln\left(x^{2}+4x\right)
x=a is a point of inflexion for a function f if:
- f' is differentiable in x=a
- f''\left(a\right)=0
- f'' changes its sign in x=a
Calculate f'\left(x\right) :
f'\left(x\right)=\left(\ln\left(x^{2}+4x\right)\right)'=\dfrac{\left(x^{2}+4x\right)'}{x^{2}+4x}=\dfrac{2x+4}{x^{2}+4x}
Calculate f''\left(x\right) :
f''\left(x\right)=\left( \dfrac{2x+4}{x^{2}+4x} \right)'=\dfrac{\left(2x+4\right)'\left(x^{2}+4x\right)-\left(2x+4\right)\left(x^{2}+4x\right)'}{\left(x^{2}+4x\right)^{2}}=\dfrac{2\left(x^{2}+4x\right)-\left(2x+4\right)\left(2x+4\right)}{\left(x^{2}+4x\right)^{2}}=\dfrac{-2x^{2}-8x-16}{\left(x^{2}+4x\right)^{2}}
Solve the equation f''\left(x\right)=0 :
\dfrac{-2x^{2}-8x-16}{\left(x^{2}+4x\right)^{2}}=0
The equation has no real solutions.
f has no inflexion points.
Determine the x-coordinates of the points of inflexion of the following function:
f\left(x\right)=\dfrac{1}{2}x^{4}+\dfrac{2}{3}x^{3}-5x+4
x=a is a point of inflexion for a function f if:
- f' is differentiable in x=a
- f''\left(a\right)=0
- f'' changes its sign in x=a
Calculate f'\left(x\right) :
f'\left(x\right)=\left(\dfrac{1}{2}x^{4}+\dfrac{2}{3}x^{3}-5x+4\right)'=2x^{3}+2x^{2}-5
Calculate f''\left(x\right) :
f''\left(x\right)=\left(2x^{3}+2x^{2}-5\right)'=6x^{2}+4x
Solve the equation f''\left(x\right)=0 :
6x^{2}+4x=0
The solutions are:
x=0 and x=-\dfrac{2}{3}
The sign of f'' is the following:
The points of inflexion for the function f are x=0 and x=-\dfrac{2}{3}.
Determine the x-coordinates of the points of inflexion of the following function:
f\left(x\right)=x^{3}\ln\left(x\right)
x=a is a point of inflexion for a function f if:
- f' is differentiable in x=a
- f''\left(a\right)=0
- f'' changes its sign in x=a
Calculate f'\left(x\right) :
f'\left(x\right)=\left(x^{3}\ln\left(x\right)\right)'=\left( x^{3} \right)'\cdot\ln\left(x\right)+\left( x^{3} \right)\cdot\left( \ln\left(x\right) \right)'=3x^{2}\ln\left(x\right)+x^{3}\cdot\dfrac{1}{x}=3x^{2}\ln\left(x\right)+x^{2}=x^{2}\left(3\ln\left(x\right)+1\right)
Calculate f''\left(x\right) :
f''\left(x\right)=\left( x^{2}\left(3\ln\left(x\right)+1\right) \right)'=\left( x^{2}\right)'\left( 3\ln\left(x\right)+1 \right)+x^{2}\left( 3\ln\left(x\right)+1 \right)'=2x\left( 3\ln\left(x\right)+1 \right)+x^{2}\cdot\dfrac{3}{x}=x\left( 6\ln\left(x\right)+5 \right)
Solve the equation f''\left(x\right)=0 for x\gt0 :
x\left( 6\ln\left(x\right)+5 \right)=0
The solution is:
x={e^{-\frac{5}{6}}}
Since f'' changes its sign in x={e^{-\frac{5}{6}}}, we conclude that x={e^{-\frac{5}{6}}} is an inflexion point.
The point of inflexion for the function f is x={e^{-\frac{5}{6}}}.
Determine the x-coordinates of the points of inflexion of the following function:
f\left(x\right)=4x^{2}-10x+\ln\left(7\right)
x=a is a point of inflexion for a function f if:
- f' is differentiable in x=a
- f''\left(a\right)=0
- f'' changes its sign in x=a
Calculate f'\left(x\right) :
f'\left(x\right)=\left(4x^{2}-10x+\ln\left(7\right)\right)'=8x-10
Calculate f''\left(x\right) :
f''\left(x\right)=\left( 8x-10 \right)'=8
Solve the equation f''\left(x\right)=0 :
8=0
The equation has no solution.
The function f has no inflexion points.