The question
If f ( x + y ) = f ( x ) f ( y ) f(x + y) = f(x)f(y) f ( x + y ) = f ( x ) f ( y ) x , y ∈ R x, y \in R x , y ∈ R f ( 5 ) = 2 f(5) = 2 f ( 5 ) = 2 f ′ ( 0 ) = 3 f'(0) = 3 f ′ ( 0 ) = 3 f ′ ( 5 ) f'(5) f ′ ( 5 )
Exam answer Show all explanations
Given f ( x + y ) = f ( x ) f ( y ) f(x + y) = f(x)f(y) f ( x + y ) = f ( x ) f ( y ) x = 0 x = 0 x = 0 y = 5 y = 5 y = 5
f ( 0 + 5 ) = f ( 0 ) f ( 5 ) ⇒ 2 = 2 f ( 0 ) ⇒ f ( 0 ) = 1 \begin{aligned} f(0 + 5) &= f(0)f(5) \\ \Rightarrow 2 &= 2f(0) \Rightarrow f(0) = 1 \end{aligned} f ( 0 + 5 ) ⇒ 2 = f ( 0 ) f ( 5 ) = 2 f ( 0 ) ⇒ f ( 0 ) = 1 Why?
The rule f ( x + y ) = f ( x ) f ( y ) f(x + y) = f(x)f(y) f ( x + y ) = f ( x ) f ( y ) x x x y y y f ( 0 ) f(0) f ( 0 ) x = 0 x = 0 x = 0 y = 5 y = 5 y = 5
That gives f ( 5 ) = f ( 0 ) f ( 5 ) f(5) = f(0)f(5) f ( 5 ) = f ( 0 ) f ( 5 ) 2 = 2 f ( 0 ) 2 = 2f(0) 2 = 2 f ( 0 ) f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1
By the definition of the derivative,
f ′ ( 5 ) = lim h → 0 f ( 5 + h ) − f ( 5 ) h = lim h → 0 f ( 5 ) f ( h ) − f ( 5 ) h = lim h → 0 2 f ( h ) − 2 h \begin{aligned} f'(5) &= \lim_{h \to 0}\frac{f(5 + h) - f(5)}{h} \\ &= \lim_{h \to 0}\frac{f(5)f(h) - f(5)}{h} \\ &= \lim_{h \to 0}\frac{2f(h) - 2}{h} \end{aligned} f ′ ( 5 ) = h → 0 lim h f ( 5 + h ) − f ( 5 ) = h → 0 lim h f ( 5 ) f ( h ) − f ( 5 ) = h → 0 lim h 2 f ( h ) − 2 Why?
The question asks for the definition (first principles), not a formula:
f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h f'(a) = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h} f ′ ( a ) = h → 0 lim h f ( a + h ) − f ( a ) with a = 5 a = 5 a = 5 f ( 5 + h ) f(5 + h) f ( 5 + h ) x = 5 x = 5 x = 5 y = h y = h y = h f ( 5 + h ) = f ( 5 ) f ( h ) f(5 + h) = f(5)f(h) f ( 5 + h ) = f ( 5 ) f ( h ) f ( 5 ) = 2 f(5) = 2 f ( 5 ) = 2
f ′ ( 5 ) = 2 lim h → 0 f ( h ) − 1 h = 2 lim h → 0 f ( 0 + h ) − f ( 0 ) h = 2 f ′ ( 0 ) = 2 ( 3 ) \begin{aligned} f'(5) &= 2\lim_{h \to 0}\frac{f(h) - 1}{h} \\ &= 2\lim_{h \to 0}\frac{f(0 + h) - f(0)}{h} \\ &= 2f'(0) = 2(3) \end{aligned} f ′ ( 5 ) = 2 h → 0 lim h f ( h ) − 1 = 2 h → 0 lim h f ( 0 + h ) − f ( 0 ) = 2 f ′ ( 0 ) = 2 ( 3 ) (using f ( 5 + h ) = f ( 5 ) f ( h ) f(5 + h) = f(5)f(h) f ( 5 + h ) = f ( 5 ) f ( h ) f ( 5 ) = 2 f(5) = 2 f ( 5 ) = 2 f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1 f ′ ( 0 ) = 3 f'(0) = 3 f ′ ( 0 ) = 3
Why?
Take the common factor 2 out of the limit. Then write 1 = f ( 0 ) 1 = f(0) 1 = f ( 0 ) h = 0 + h h = 0 + h h = 0 + h
2 f ( h ) − 2 h = 2 ( f ( h ) − 1 ) h = 2 ⋅ f ( h ) − 1 h \begin{aligned} \frac{2f(h) - 2}{h} &= \frac{2\big(f(h) - 1\big)}{h} \\ &= 2 \cdot \frac{f(h) - 1}{h} \end{aligned} h 2 f ( h ) − 2 = h 2 ( f ( h ) − 1 ) = 2 ⋅ h f ( h ) − 1 The limit left is exactly the definition of f ′ ( 0 ) f'(0) f ′ ( 0 )
∴ f ′ ( 5 ) = 6 \therefore\ f'(5) = 6 ∴ f ′ ( 5 ) = 6
Strictly, the given data cannot all hold: f ( x + y ) = f ( x ) f ( y ) f(x + y) = f(x)f(y) f ( x + y ) = f ( x ) f ( y ) f ′ ( 0 ) = 3 f'(0) = 3 f ′ ( 0 ) = 3 f ( x ) = e 3 x f(x) = e^{3x} f ( x ) = e 3 x f ( 5 ) f(5) f ( 5 ) e 15 e^{15} e 15 f ′ ( 5 ) = f ( 5 ) f ′ ( 0 ) = 6 f'(5) = f(5)f'(0) = 6 f ′ ( 5 ) = f ( 5 ) f ′ ( 0 ) = 6