Let h(x) = e^(kx) + f(x),

1 answer:

3 0

$\bf h(x)=e^{kx}+f(x)\implies \cfrac{dh}{dx}=ke^{kx}+f'(x)
\\\\\\
\begin{cases}
x=0\\
f'(0)=-8
\end{cases} 
h'(0)=ke^{k(0)}+(-8)\implies h'(0)=k-8$

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Vaselesa [24]

So when you're solving for x here and distribute 8 on the right side, it would look like this:

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But we can't! Since both sides will cancel out to zero before x can be solved.

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let us denote the diagonal of rectangle by D and the sides of rectangle be denoted by X=12 and Y=9

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