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f^(prime)(x)=varphi^(prime)(x)=f(x) for ...

`f^(prime)(x)=varphi^(prime)(x)=f(x)` for all `xdot` Also, `f(3)=5a n df^(prime)(3)=4.` Then the value of `[f(10)]^2` is _____

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`(d)/(dx){[(f(x)]^(2)-[phi(x)]^(2)}`
`=2[f(x)cdotf'(x)-phi(x)cdotphi(x)]`
`=2[f(x)cdotphi(x)-phi(x)cdotf(x)][becausef'(x)=phi(x) and phi'(x)=f(x)]=0`
`"or "[f(x)]^(2)-[phi(x)]^(2)`= constant
`therefore" "[f(10)]^(2)-[phi(10)]^(2)=[f(3)]^(2)-[phi(3)]^(2)=[f(3)]^(2)-[f'(3)]^(2)`
`=25-16=9.`
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