Let f be the function defined as follows:
f:x\longmapsto 2x^2-4x
Determine whether f is increasing or decreasing between 1 and 5.
Recall that:
- If f'\left(x\right) \gt 0 when x\in\left( a,b \right), then f is increasing on the interval \left( a,b \right).
- If f'\left(x\right) \lt 0 when x\in\left( a,b \right), then f is decreasing on the interval \left( a,b \right).
In our problem:
- f\left(x\right)=2x^{2}-4x
- The interval is \left[ 1{,}5 \right]
Calculate the derivative:
f'\left(x\right)=\left(2x^{2}-4x\right)'=4x-4
In order to find the sign of f' on the interval \left[ 1{,}5 \right], first we solve the equation f'\left(x\right)=0 :
f'\left(x\right)=0
4x-4=0
x=1
We find that for any number greater than x=1, f'\left(x\right) \gt 0.
f is increasing on \left[1{,}5\right]
Let f be the function defined as follows:
f:x\longmapsto x^{3}-4x^{2}+5x-7
Determine whether f is increasing or decreasing.
Recall that:
- If f'\left(x\right) \gt 0 when x\in\left( a,b \right), then f is increasing on the interval \left( a,b \right).
- If f'\left(x\right) \lt 0 when x\in\left( a,b \right), then f is decreasing on the interval \left( a,b \right).
In our problem:
f\left(x\right)=x^{3}-4x^{2}+5x-7
Calculate the derivative:
f'\left(x\right)=\left(x^{3}-4x^{2}+5x-7\right)'=3x^{2}-8x+5
In order to find the sign of f', first solve the equation f'\left(x\right)=0 :
f'\left(x\right)=0
3x^{2}-8x+5=0
\left(x-1\right)\left(3x-5\right)=0
The solutions of the equation are:
x=1 and x=\dfrac{5}{3}
The sign of the derivative is as follows:
f is increasing when x\in\left(-\infty,1\right)\cup\left(\dfrac{5}{3},\infty\right) and decreasing when x\in\left(1,\dfrac{5}{3}\right).
Let f be the function defined as follows:
f:x\longmapsto x^{2}\ln\left(x\right)
Determine whether f is increasing or decreasing on \left(0,\infty\right).
Recall that:
- If f'\left(x\right) \gt 0 when x\in\left( a,b \right), then f is increasing on the interval \left( a,b \right).
- If f'\left(x\right) \lt 0 when x\in\left( a,b \right), then f is decreasing on the interval \left( a,b \right).
In our problem:
f\left(x\right)=x^{2}\ln\left(x\right)
Calculate the derivative:
f'\left(x\right)=\left(x^{2}\ln\left(x\right)\right)'=\left(x^{2}\right)'\ln\left(x\right)+x^{2}\left(\ln\left(x\right)\right)'=2x\ln\left(x\right)+x^{2}\cdot\dfrac{1}{x}=2x\ln\left(x\right)+x=x\left(2\ln\left(x\right)+1\right)
In order to find the sign of f', first solve the equation f'\left(x\right)=0 :
f'\left(x\right)=0
x\left(2\ln\left(x\right)+1\right)=0
The solutions of the equation are:
x=0
and
2\ln\left(x\right)+1=0
\ln\left(x\right)=-\dfrac{1}{2}
x=e^{-\frac{1}{2}}
x=\dfrac{1}{\sqrt{e}}
x=\dfrac{\sqrt{e}}{e}
The sign of the derivative is as follows:
f is decreasing when x\in\left(0,\dfrac{\sqrt{e}}{e}\right) and increasing when x\in\left(\dfrac{\sqrt{e}}{e},\infty\right).
Let f be the function defined as follows:
f:x\longmapsto \dfrac{x^{2}}{x-1}
Determine whether f is increasing or decreasing on \left(-\infty,1\right)\cup\left(1,\infty\right).
Recall that:
- If f'\left(x\right) \gt 0 when x\in\left( a,b \right), then f is increasing on the interval \left( a,b \right).
- If f'\left(x\right) \lt 0 when x\in\left( a,b \right), then f is decreasing on the interval \left( a,b \right).
In our problem:
f\left(x\right)=\dfrac{x^{2}}{x-1}
Calculate the derivative:
f'\left(x\right)=\left(\dfrac{x^{2}}{x-1}\right)'=\dfrac{\left(x^{2}\right)'\left(x-1\right)-x^{2}\left(x-1\right)'}{\left(x-1\right)^{2}}=\dfrac{2x\left(x-1\right)-x^{2}}{\left(x-1\right)^{2}}=\dfrac{x^{2}-2x}{\left(x-1\right)^{2}}
In order to find the sign of f', first solve the equation f'\left(x\right)=0 :
f'\left(x\right)=0
\dfrac{x^{2}-2x}{\left(x-1\right)^{2}}=0
The solutions of the equation are:
x=0 and x=2
The sign of the derivative is as follows:
f is decreasing when x\in\left(0{,}1\right)\cup\left(1{,}2\right) and increasing when x\in\left(-\infty,0\right)\cup\left(2,\infty\right).
Let f be the function defined as follows:
f:x\longmapsto x^{4}e^{2x}
Determine whether f is increasing or decreasing.
Recall that:
- If f'\left(x\right) \gt 0 when x\in\left( a,b \right), then f is increasing on the interval \left( a,b \right).
- If f'\left(x\right) \lt 0 when x\in\left( a,b \right), then f is decreasing on the interval \left( a,b \right).
In our problem:
f\left(x\right)=x^{4}e^{2x}
Calculate the derivative:
f'\left(x\right)=\left(x^{4}e^{2x}\right)'=\left(x^{4}\right)'e^{2x}+x^{4}\left(e^{2x}\right)'=4x^{3}e^{2x}+x^{4}e^{2x}\cdot2=2x^{3}e^{2x}\left(x+2\right)
In order to find the sign of f', first solve the equation f'\left(x\right)=0 :
f'\left(x\right)=0
2x^{3}e^{2x}\left(x+2\right)=0
The solutions of the equation are:
- x=0
- x=-2
The sign of the derivative is as follows:
f is increasing when x\in\left(-\infty,-2\right)\cup\left(0,\infty\right) and decreasing when x\in\left(-2{,}0\right).
Let f be the function defined as follows:
f:x\longmapsto \sqrt{x^{2}+4x+5}
Determine whether f is increasing or decreasing.
Recall that:
- If f'\left(x\right) \gt 0 when x\in\left( a,b \right), then f is increasing on the interval \left( a,b \right).
- If f'\left(x\right) \lt 0 when x\in\left( a,b \right), then f is decreasing on the interval \left( a,b \right).
In our problem:
f\left(x\right)=\sqrt{x^{2}+4x+5}
Calculate the derivative:
f'\left(x\right)=\left(\sqrt{x^{2}+4x+5}\right)'=\dfrac{\left(x^{2}+4x+5\right)'}{2\sqrt{x^{2}+4x+5}}=\dfrac{2x+4}{2\sqrt{x^{2}+4x+5}}=\dfrac{x+2}{\sqrt{x^{2}+4x+5}}
In order to find the sign of f', first solve the equation f'\left(x\right)=0 :
f'\left(x\right)=0
\dfrac{x+2}{\sqrt{x^{2}+4x+5}}=0
The solution of the equation is:
x=-2
The sign of the derivative is as follows:
f is decreasing when x\in\left(-\infty,-2\right) and increasing when x\in\left(-2,\infty\right).
Let f be the function defined as follows:
f:x\longmapsto \dfrac{x+3}{2x-4}
Determine whether f is increasing or decreasing on \left(-\infty,2\right)\cup\left(2,\infty\right).
Recall that:
- If f'\left(x\right) \gt 0 when x\in\left( a,b \right), then f is increasing on the interval \left( a,b \right).
- If f'\left(x\right) \lt 0 when x\in\left( a,b \right), then f is decreasing on the interval \left( a,b \right).
In our problem:
f\left(x\right)=\dfrac{x+3}{2x-4}
Calculate the derivative:
f'\left(x\right)=\left(\dfrac{x+3}{2x-4}\right)'=\dfrac{\left(x+3\right)'\left(2x-4\right)-\left(x+3\right)\left(2x-4\right)'}{\left(x-4\right)^{2}}=\dfrac{1\cdot\left(2x-4\right)-\left(x+3\right)\cdot2}{\left(x-4\right)^{2}}=\dfrac{-10}{\left(x-4\right)^{2}}
In order to find the sign of f', first solve the equation f'\left(x\right)=0 :
f'\left(x\right)=0
\dfrac{-10}{\left(x-4\right)^{2}}=0
The equation has no solution.
We conclude that f is decreasing when x\in\left(-\infty,2\right)\cup\left(2,\infty\right).
f is decreasing when x\in\left(-\infty,2\right)\cup\left(2,\infty\right).