# Given x^5 if x<0 and 2x + 1 if x >= 0, how do you find f(0)?

Aug 7, 2016

$\implies f \left(0\right) = 0 + 1 = 1$

#### Explanation:

I think they call this a 'stepwise solution'

$f \left(0\right)$ means that you assign the value of 0 to $x$

Condition 1: if $x < 0$ then you apply the calculation ${x}^{5}$

Condition 2: if $x \ge 0$ then you apply the calculation $2 x + 1$
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We are given the value of $x$ as 0

Condition 1 is that $x$ must be less than 0 so as $x = 0$ this equation does not apply.

Condition 2 is that $x$ can be 0 or grater than 0. So as $x = 0$ this equation is the one we use.

$\implies f \left(0\right) \to 2 x + 1$

$\implies f \left(0\right) = 2 \left(0\right) + 1$

$\implies f \left(0\right) = 0 + 1 = 1$