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Sagot :
Answer:
Given function is,
A) To find values of all local maxima of f.
[tex]f\mleft(x\mright)=x^3-9x^2+10[/tex]Consider the derivative of f(x), we get,
[tex]f\mleft(x\mright)=x^3-9x^2+10[/tex][tex]f^{\prime}\left(x\right)[/tex]we get,
[tex]f^{\prime}\mleft(x\mright)=3x^2-18x^[/tex]Also, let f'(x) be zero, we get (f'(x)=0),
[tex]3x^2-18x=0[/tex]Simplifing we get,
[tex]3x\left(x-6\right)=0[/tex]we get,
[tex]x=0\text{ or x=6}[/tex]To find the local maximum,
if f'(x-c)>0 anf f'(x+c)<0, then the x is local maximum
f f'(x-c)<0 anf f'(x+c)>0, then the x is local minimum
For x=0
Consider x-c as -1 (x-c=-1), we get
[tex]f^{\prime}(-1)=3\left(-1\right)^2-18\left(-1\right)[/tex][tex]f^{\prime}(-1)=21>0[/tex]Consider x+c as 1, (x+c=1), we get
[tex]f^{\prime}\mleft(1\mright)=3\left(1\right)^2-18\left(1\right)^[/tex][tex]f^{\prime}\mleft(1\mright)=-15<0[/tex]Since f'(-1)>0 anf f'(1)<0, then the 0 is local maximum.
x value of local maximum = 0
Answer is: x value of local maximum = 0
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