A derivative problem

Calculus Level 5

Let f k ( x ) = 0 ; w h e n x < k f_k(x) =0 ; when \quad x < k and f k ( x ) = 1 ; w h e n x ≥ k f_k(x)=1 ; when \quad x \ge k . The function 'T' is defined to be T k ( x ) = d d x ( f k ( x ) ) T_k(x) = \dfrac{d}{dx} (f_k(x)) Suppose d 2 g ( x ) d x 2 = T 1 ( x ) + T 2 ( x ) \dfrac{d^2 g(x)}{dx^2} = T_1(x) + T_2 (x) and g ( 0 ) = g ′ ( 0 ) = 0 g(0) = g'(0) = 0 Find g ( 5 ) \displaystyle{g(5)}


The answer is 7.

Abhishek Singh by Abhishek Singh , 24, India

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1 solution

Abhishek Singh
Nov 4, 2014

g ′ ( x ) = ∫ ( T 1 ( x ) + T 2 ( x ) ) d x = f 1 ( x ) + f 2 ( x ) + c g'(x)= \int (T_1(x) + T_2(x)) dx = f_1(x) + f_2(x) +c and g ′ ( 0 ) = 0 ⇒ c = 0 \displaystyle{ g'(0) = 0 \Rightarrow c=0} So, g ′ ( x ) = f 1 ( x ) + f 2 ( x ) \displaystyle{g'(x) = f_1(x) + f_2(x)} So g ( 5 ) \displaystyle{g(5)} is the area of the two rectangles from x=1 to x=5 and x=2 to x=5 Which comes out to be 7 \displaystyle{\boxed{7}}

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