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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`− [ϕ(10)]^2is _____

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To solve the problem step by step, we start with the given information: 1. **Given Equations**: - \( f'(x) = \varphi'(x) = f(x) \) for all \( x \) - \( f(3) = 5 \) - \( f'(3) = 4 \) 2. **Objective**: ...
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CENGAGE ENGLISH-DIFFERENTIATION-Numerical Value Type
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