Functions and more functions

Calculus Level 3

Let us define 2 functions $f(x)$ and $g(x)$ such that

$f ^{\prime \prime} (x) = - f(x)$ and,

$g(x) = f^{ \prime }(x)$

Now lets define a new function $H(x)$ such that

$H(x) = \left( f \left(\dfrac x2\right) \right)^2 + \left( g\left(\dfrac x2\right) \right)^2$

It is given that $H(e) = \pi$ . Then find value of $H(\pi)$ upto 2 decimal places

Notation :

$f ^{\prime} (x)$ refers to $\displaystyle \frac{df(x)}{dx}$ .
- $f ^{\prime \prime} (x)$ refers to $\displaystyle \frac{d^2f(x)}{dx^2}$ .

Try more here

The answer is 3.14.

1 solution

Neelesh Vij
Feb 15, 2016

Firstly lets find a relation between the functions,

We are given that $g(x) = f^{\prime} (x)$ , Differentiating w.r.t. $x$

$\Rightarrow g^{\prime}(x) = f^{\prime \prime}(x)$

$\Rrightarrow g^{\prime}(x) = -f(x)$

Differentiating again w.r.t. $x$ we get

$\Rightarrow g^{\prime \prime}(x) = -g(x)$

Lets looks at the function $H(x)$

$H^{\prime}(x) = 2f \left( \dfrac x2 \right) \times 2f^{\prime} \left( \dfrac x2 \right) + 2g \left( \dfrac x2 \right) \times 2g^{\prime} \left( \dfrac x2 \right)$

Now putting values of $g^{\prime}(x) and f^{\prime}(x)$

$\Rightarrow H^{\prime}(x) = 2f \left( \dfrac x2 \right) \times g \left( \dfrac x2 \right) - 2f \left( \dfrac x2 \right) \times g \left( \dfrac x2 \right)$

$\therefore H^{\prime}(x) = 0$ , So $H(x)$ is a constant function

$\Rightarrow H(e) = H(\pi) = \pi = \boxed{3.14}$

Looks great. I might add "to two decimal places" inside the problem statement so that people know you're expecting a rounded answer (even if the answer blanks says you can use decimals)

Andrew Ellinor - 5 years, 3 months ago

Functions and more functions

The answer is 3.14.

1 solution

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