We need the value of λ \lambda λ f ′ ′ ( 0 ) f''(0) f ′′ ( 0 )
Given
f ( x ) = { x 5 sin ( 1 x ) + 5 x 2 , x < 0 0 , x = 0 x 5 cos ( 1 x ) + λ x 2 , x > 0 f(x)=
\begin{cases}
x^5\sin\left(\frac1x\right)+5x^2, & x<0 \\
0, & x=0 \\
x^5\cos\left(\frac1x\right)+\lambda x^2, & x>0
\end{cases} f ( x ) = ⎩ ⎨ ⎧ x 5 sin ( x 1 ) + 5 x 2 , 0 , x 5 cos ( x 1 ) + λ x 2 , x < 0 x = 0 x > 0 To ensure f ′ ′ ( 0 ) f''(0) f ′′ ( 0 ) f ′ ( 0 ) f'(0) f ′ ( 0 )
f ′ ′ ( 0 ) = lim h → 0 f ′ ( h ) − f ′ ( 0 ) h f''(0)=\lim_{h\to 0}\frac{f'(h)-f'(0)}{h} f ′′ ( 0 ) = h → 0 lim h f ′ ( h ) − f ′ ( 0 ) must exist.
First compute f ′ ( 0 ) f'(0) f ′ ( 0 )
By definition,
f ′ ( 0 ) = lim h → 0 f ( h ) − f ( 0 ) h = lim h → 0 f ( h ) h f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{f(h)}{h} f ′ ( 0 ) = h → 0 lim h f ( h ) − f ( 0 ) = h → 0 lim h f ( h ) since f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0
For h < 0 h<0 h < 0
f ( h ) h = h 4 sin ( 1 h ) + 5 h \frac{f(h)}{h}=h^4\sin\left(\frac1h\right)+5h h f ( h ) = h 4 sin ( h 1 ) + 5 h For h > 0 h>0 h > 0
f ( h ) h = h 4 cos ( 1 h ) + λ h \frac{f(h)}{h}=h^4\cos\left(\frac1h\right)+\lambda h h f ( h ) = h 4 cos ( h 1 ) + λh As h → 0 h\to 0 h → 0 0 0 0
Hence,
f ′ ( 0 ) = 0. f'(0)=0. f ′ ( 0 ) = 0.
Now compute f ′ ( x ) f'(x) f ′ ( x ) x ≠ 0 x\neq 0 x = 0
For x < 0 x<0 x < 0
f ( x ) = x 5 sin ( 1 x ) + 5 x 2 f(x)=x^5\sin\left(\frac1x\right)+5x^2 f ( x ) = x 5 sin ( x 1 ) + 5 x 2 Differentiate:
d d x ( x 5 sin ( 1 x ) ) = 5 x 4 sin ( 1 x ) + x 5 cos ( 1 x ) ( − 1 x 2 ) \frac{d}{dx}\left(x^5\sin\left(\frac1x\right)\right)=5x^4\sin\left(\frac1x\right)+x^5\cos\left(\frac1x\right)\left(-\frac1{x^2}\right) d x d ( x 5 sin ( x 1 ) ) = 5 x 4 sin ( x 1 ) + x 5 cos ( x 1 ) ( − x 2 1 ) so
d d x ( x 5 sin ( 1 x ) ) = 5 x 4 sin ( 1 x ) − x 3 cos ( 1 x ) . \frac{d}{dx}\left(x^5\sin\left(\frac1x\right)\right)=5x^4\sin\left(\frac1x\right)-x^3\cos\left(\frac1x\right). d x d ( x 5 sin ( x 1 ) ) = 5 x 4 sin ( x 1 ) − x 3 cos ( x 1 ) . Also,
d d x ( 5 x 2 ) = 10 x . \frac{d}{dx}(5x^2)=10x. d x d ( 5 x 2 ) = 10 x . Thus,
f ′ ( x ) = 5 x 4 sin ( 1 x ) − x 3 cos ( 1 x ) + 10 x , x < 0. f'(x)=5x^4\sin\left(\frac1x\right)-x^3\cos\left(\frac1x\right)+10x, \qquad x<0. f ′ ( x ) = 5 x 4 sin ( x 1 ) − x 3 cos ( x 1 ) + 10 x , x < 0. For x > 0 x>0 x > 0
f ( x ) = x 5 cos ( 1 x ) + λ x 2 f(x)=x^5\cos\left(\frac1x\right)+\lambda x^2 f ( x ) = x 5 cos ( x 1 ) + λ x 2 Differentiate:
d d x ( x 5 cos ( 1 x ) ) = 5 x 4 cos ( 1 x ) + x 5 ( − sin ( 1 x ) ) ( − 1 x 2 ) \frac{d}{dx}\left(x^5\cos\left(\frac1x\right)\right)=5x^4\cos\left(\frac1x\right)+x^5\left(-\sin\left(\frac1x\right)\right)\left(-\frac1{x^2}\right) d x d ( x 5 cos ( x 1 ) ) = 5 x 4 cos ( x 1 ) + x 5 ( − sin ( x 1 ) ) ( − x 2 1 ) Hence,
d d x ( x 5 cos ( 1 x ) ) = 5 x 4 cos ( 1 x ) + x 3 sin ( 1 x ) . \frac{d}{dx}\left(x^5\cos\left(\frac1x\right)\right)=5x^4\cos\left(\frac1x\right)+x^3\sin\left(\frac1x\right). d x d ( x 5 cos ( x 1 ) ) = 5 x 4 cos ( x 1 ) + x 3 sin ( x 1 ) . Also,
d d x ( λ x 2 ) = 2 λ x . \frac{d}{dx}(\lambda x^2)=2\lambda x. d x d ( λ x 2 ) = 2 λ x . Thus,
f ′ ( x ) = 5 x 4 cos ( 1 x ) + x 3 sin ( 1 x ) + 2 λ x , x > 0. f'(x)=5x^4\cos\left(\frac1x\right)+x^3\sin\left(\frac1x\right)+2\lambda x, \qquad x>0. f ′ ( x ) = 5 x 4 cos ( x 1 ) + x 3 sin ( x 1 ) + 2 λ x , x > 0.
Now check existence of f ′ ′ ( 0 ) f''(0) f ′′ ( 0 )
Since f ′ ( 0 ) = 0 f'(0)=0 f ′ ( 0 ) = 0
f ′ ′ ( 0 ) = lim h → 0 f ′ ( h ) h . f''(0)=\lim_{h\to 0}\frac{f'(h)}{h}. f ′′ ( 0 ) = h → 0 lim h f ′ ( h ) . Left-hand limit (h → 0 − h\to 0^- h → 0 −
Using the expression for h < 0 h<0 h < 0
f ′ ( h ) h = 5 h 3 sin ( 1 h ) − h 2 cos ( 1 h ) + 10. \frac{f'(h)}{h}=5h^3\sin\left(\frac1h\right)-h^2\cos\left(\frac1h\right)+10. h f ′ ( h ) = 5 h 3 sin ( h 1 ) − h 2 cos ( h 1 ) + 10. As h → 0 − h\to 0^- h → 0 −
5 h 3 sin ( 1 h ) → 0 , − h 2 cos ( 1 h ) → 0. 5h^3\sin\left(\frac1h\right)\to 0, \qquad -h^2\cos\left(\frac1h\right)\to 0. 5 h 3 sin ( h 1 ) → 0 , − h 2 cos ( h 1 ) → 0. So,
lim h → 0 − f ′ ( h ) h = 10. \lim_{h\to 0^-}\frac{f'(h)}{h}=10. h → 0 − lim h f ′ ( h ) = 10. Right-hand limit (h → 0 + h\to 0^+ h → 0 +
Using the expression for h > 0 h>0 h > 0
f ′ ( h ) h = 5 h 3 cos ( 1 h ) + h 2 sin ( 1 h ) + 2 λ . \frac{f'(h)}{h}=5h^3\cos\left(\frac1h\right)+h^2\sin\left(\frac1h\right)+2\lambda. h f ′ ( h ) = 5 h 3 cos ( h 1 ) + h 2 sin ( h 1 ) + 2 λ . As h → 0 + h\to 0^+ h → 0 +
5 h 3 cos ( 1 h ) → 0 , h 2 sin ( 1 h ) → 0. 5h^3\cos\left(\frac1h\right)\to 0, \qquad h^2\sin\left(\frac1h\right)\to 0. 5 h 3 cos ( h 1 ) → 0 , h 2 sin ( h 1 ) → 0. So,
lim h → 0 + f ′ ( h ) h = 2 λ . \lim_{h\to 0^+}\frac{f'(h)}{h}=2\lambda. h → 0 + lim h f ′ ( h ) = 2 λ . For f ′ ′ ( 0 ) f''(0) f ′′ ( 0 )
10 = 2 λ . 10=2\lambda. 10 = 2 λ . Therefore,
λ = 5. \lambda=5. λ = 5.
Final answer:
5 \boxed{5} 5 This matches the stored correct answer.