Question

Calculus,

https://www.dropbox.com/s/uirpmtxt6f2mhid/Screenshot%202016-01-07%2020.40.38.png?dl=0


for (34) is there a maximum at x=4 ? Or am I wrong ?

Answer 1

\[f \left( x \right)=x^4-4x^3+4x^2\]
\[f \prime \left( x \right)=4x^3-12x^2+8x=4x \left( x^2-3x+2 \right)\]
\[=4x \left\{ x^2-2x-x+2 \right\}=4x \left\{ x \left( x-2 \right)-1\left( x-2 \right) \right\}=4x \left( x-2 \right)\left( x-1 \right)\]
f(x)=0 gives x=0,1,2
\[f \prime \prime \left( x \right)=12x^2-24x+8=12(x^2-2x+1-1)+8=12\left( x-1 \right)^2-4\]
at x=0 \[f \prime \prime \left( x \right)=8>0,\]
f(x) has relative minima at x=0
at x=1
\[f \prime \prime \left( x \right)=-4<0\]
f(x) has relative maxima at x =1
at x=2
\[f \prime \prime \left( x \right)=8>0\]
f(x)has relative minima at x=2

Answer 2

34.
\[f \left( x \right)=x \left( x-4 \right)^3\]
\[f \prime \left( x \right)=x*3\left( x-4 \right)^2+1*\left( x-4 \right)^3=\left( x-4 \right)^2\left\{ 3x+x-4 \right\}\]
\[=4\left( x-4 \right)^2\left( x-1 \right)\]
\[f \prime \left( x \right)=0,x=4,1\]
when x<1 (slightly)
x-1<0
\[f \prime \left( x \right)=\left( + \right)\left( - \right)= -ve\]
when x>1
x-1>0
\[f \prime \left( x \right)=\left( + \right)\left( + \right)=+ve\]
\[f \prime \left( x \right)~changes~sign~from~-ve~\to~+ve~as~x~passes~through~1\]
there is relative minima at x=1
at x=4,\[f \prime \left( x \right) ~does~\not~changes~sign~at~x=4\]
there is a cusp at x=4