Given that f''(x)=\cos (x), f' \frac {\pi}{2}=14, f \frac {\pi}{2}=12, find f'(x) and f(x).

Question:

Given that {eq}f''(x)=\cos (x), f' (\frac {\pi}{2})=14, f (\frac {\pi}{2})=12 {/eq}, find {eq}f'(x) {/eq} and {eq}f(x) {/eq}.

Constant of Anti-derivatives:


When we apply the general rule of Anti-derivative and derivatives shown below, we get a constant of integration or anti-derivatives. For that, we'll use the given conditions (The values of the derivative function at a point) in the obtained result of anti-derivatives for the required value of constant.

{eq}\displaystyle \int g'(t)\ dt=g(t)+C {/eq}, where,

  • C is the constant of integration.

Answer and Explanation:


Given information:

The second derivative with trig-function and initial conditions are:

{eq}f''(x)=\cos (x)\\ f' \left (\frac {\pi}{2}\right )=14\\ f \left (\frac {\pi}{2}\right)=12 {/eq}

The anti-derivative of the given derivative function is:

{eq}\begin{align*} \displaystyle \int f''(x)\ dx& =\int \cos (x)\ dx\\ \displaystyle f'(x)&= \sin (x)+C..................(1)\\ \end{align*} {/eq}

Accordiing to the initial condition {eq}f' \left(\frac {\pi}{2}\right)=14 {/eq} and the above-obtained expression, we have:

{eq}\begin{align*} \displaystyle f'\left (\frac {\pi}{2}\right )&= \sin \left (\frac {\pi}{2}\right )+C\\ \displaystyle 14&= 1+C\\ \displaystyle C&= 14-1\\ \displaystyle C&= 13\\ \end{align*} {/eq}


Plugging obtained value of C in equation (1) and simplifying it, we get:

{eq}\begin{align*} \displaystyle f'(x)= \sin (x)+13 \end{align*} {/eq}

The anti-derivative of both sides of the above expression is:

{eq}\begin{align*} \displaystyle \int f'(x)\ dx&=\displaystyle \int (\sin (x)+13)\ dx\\ \displaystyle f(x)&=\displaystyle \int \sin (x)\ dx+\int 13\ dx\\ &=\displaystyle -\cos (x)+13x+D\\ \end{align*} {/eq}

Where,

  • D is the constant of integration.


Plugging {eq}f \left (\frac {\pi}{2}\right)=12 {/eq} in the above expression and simplifying it for D, we get:

{eq}\begin{align*} \displaystyle f\left (\frac {\pi}{2}\right)&=\displaystyle -\cos \left (\frac {\pi}{2}\right)+13\left (\frac {\pi}{2}\right)+D\\ \displaystyle 12&=\displaystyle -0+\frac {13\pi}{2}+D\\ \displaystyle D&=\displaystyle 12-\frac {13\pi}{2}\\ \end{align*} {/eq}

Therefore, the solution of the given derivative expression is:

{eq}\boxed{\displaystyle f(x)=\displaystyle -\cos (x)+13x+12-\frac {13\pi}{2}} {/eq}


Learn more about this topic:

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Anti-Derivatives: Calculating Indefinite Integrals of Polynomials

from Math 104: Calculus

Chapter 13 / Lesson 2
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