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Sagot :
To solve this problem, we need to determine [tex]\( f(0) \)[/tex], [tex]\( f'(0) \)[/tex], and [tex]\( f'(x) \)[/tex] based on the given functional equations and limit condition.
### Step 1: Find [tex]\( f(0) \)[/tex]
We start by evaluating the functional equation for [tex]\( x = 0 \)[/tex].
[tex]\[ f(x + 0) = f(x) + f(0) + 0^2 \cdot y + 0 \cdot y^2 \][/tex]
[tex]\[ f(x) = f(x) + f(0) \][/tex]
Since [tex]\( f(x) = f(x) + f(0) \)[/tex] must hold for all [tex]\( x \)[/tex], it follows that:
[tex]\[ f(0) = 0 \][/tex]
### Step 2: Find [tex]\( f'(0) \)[/tex]
Given the limit condition:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{x} = 1 \][/tex]
This implies the derivative of [tex]\( f \)[/tex] at 0, denoted [tex]\( f'(0) \)[/tex], is given by:
[tex]\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} = 1 \][/tex]
Therefore:
[tex]\[ f'(0) = 1 \][/tex]
### Step 3: Find [tex]\( f'(x) \)[/tex]
To find the derivative [tex]\( f'(x) \)[/tex], we differentiate both sides of the functional equation [tex]\( f(x + y) = f(x) + f(y) + x^2 y + x y^2 \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[ \frac{d}{dx}\left[ f(x + y) \right] = \frac{d}{dx}\left[ f(x) + f(y) + x^2 y + x y^2 \right] \][/tex]
[tex]\[ f'(x + y) = f'(x) + \frac{d}{dx}[x^2 y] + \frac{d}{dx}[x y^2] \][/tex]
[tex]\[ f'(x + y) = f'(x) + 2xy + y^2 \][/tex]
Next, let [tex]\( y = 0 \)[/tex] in the expression:
[tex]\[ f'(x + 0) = f'(x) + 2x \cdot 0 + 0^2 \][/tex]
[tex]\[ f'(x) = f'(x) + 0 \][/tex]
This does not give new information directly. Instead, we now differentiate the limit expression:
Since the original functional equation, when differentiated with [tex]\( y \neq 0 \)[/tex], suggests the general form of [tex]\( f'(x) \)[/tex]. Consider starting with the limit:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{x} = 1 \][/tex]
We know [tex]\( f(x) \approx x \)[/tex] as [tex]\( x \to 0 \)[/tex], suggesting [tex]\( f(x) = x + h(x) \)[/tex], where [tex]\( h(x) \)[/tex] is such that [tex]\( \lim_{x \to 0} \frac{h(x)}{x} = 0 \)[/tex]. Differentiating directly:
[tex]\[ f'(x) = 1 + h'(x) \][/tex]
Given [tex]\( \frac{d}{dx}[2xy + y^2] = 2xy + y^2 \)[/tex],
comparing with the equation suggests:
Thus from above direct differentiation,
Thus,
[tex]\[ f'(x + y) = f'(x) + 2xy + y^2 \][/tex]
Thus functional equality holds,
Thus [tex]\(f'(x) = x + 1\)[/tex],
Finally:
[tex]\[ f'(x)= x+1 \][/tex]
### Final Results:
- [tex]\( f(0) = 0 \)[/tex]
- [tex]\( f'(0) = 1 \)[/tex]
- [tex]\( f'(x) = x + 1 \)[/tex]
### Step 1: Find [tex]\( f(0) \)[/tex]
We start by evaluating the functional equation for [tex]\( x = 0 \)[/tex].
[tex]\[ f(x + 0) = f(x) + f(0) + 0^2 \cdot y + 0 \cdot y^2 \][/tex]
[tex]\[ f(x) = f(x) + f(0) \][/tex]
Since [tex]\( f(x) = f(x) + f(0) \)[/tex] must hold for all [tex]\( x \)[/tex], it follows that:
[tex]\[ f(0) = 0 \][/tex]
### Step 2: Find [tex]\( f'(0) \)[/tex]
Given the limit condition:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{x} = 1 \][/tex]
This implies the derivative of [tex]\( f \)[/tex] at 0, denoted [tex]\( f'(0) \)[/tex], is given by:
[tex]\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} = 1 \][/tex]
Therefore:
[tex]\[ f'(0) = 1 \][/tex]
### Step 3: Find [tex]\( f'(x) \)[/tex]
To find the derivative [tex]\( f'(x) \)[/tex], we differentiate both sides of the functional equation [tex]\( f(x + y) = f(x) + f(y) + x^2 y + x y^2 \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[ \frac{d}{dx}\left[ f(x + y) \right] = \frac{d}{dx}\left[ f(x) + f(y) + x^2 y + x y^2 \right] \][/tex]
[tex]\[ f'(x + y) = f'(x) + \frac{d}{dx}[x^2 y] + \frac{d}{dx}[x y^2] \][/tex]
[tex]\[ f'(x + y) = f'(x) + 2xy + y^2 \][/tex]
Next, let [tex]\( y = 0 \)[/tex] in the expression:
[tex]\[ f'(x + 0) = f'(x) + 2x \cdot 0 + 0^2 \][/tex]
[tex]\[ f'(x) = f'(x) + 0 \][/tex]
This does not give new information directly. Instead, we now differentiate the limit expression:
Since the original functional equation, when differentiated with [tex]\( y \neq 0 \)[/tex], suggests the general form of [tex]\( f'(x) \)[/tex]. Consider starting with the limit:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{x} = 1 \][/tex]
We know [tex]\( f(x) \approx x \)[/tex] as [tex]\( x \to 0 \)[/tex], suggesting [tex]\( f(x) = x + h(x) \)[/tex], where [tex]\( h(x) \)[/tex] is such that [tex]\( \lim_{x \to 0} \frac{h(x)}{x} = 0 \)[/tex]. Differentiating directly:
[tex]\[ f'(x) = 1 + h'(x) \][/tex]
Given [tex]\( \frac{d}{dx}[2xy + y^2] = 2xy + y^2 \)[/tex],
comparing with the equation suggests:
Thus from above direct differentiation,
Thus,
[tex]\[ f'(x + y) = f'(x) + 2xy + y^2 \][/tex]
Thus functional equality holds,
Thus [tex]\(f'(x) = x + 1\)[/tex],
Finally:
[tex]\[ f'(x)= x+1 \][/tex]
### Final Results:
- [tex]\( f(0) = 0 \)[/tex]
- [tex]\( f'(0) = 1 \)[/tex]
- [tex]\( f'(x) = x + 1 \)[/tex]
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