Let f be a differentiable function satisfying f(x)=1-2x+ _0^xe^(x-t)f(t), dt, x R and let …

Question

Let $f$ be a differentiable function satisfying $f(x)=1-2x+\int _{0}^{x}{e}^{(x-t)}f(t)\, dt$, $x\in R$ and let
 $g(x)=\int _{0}^{x}(f(t)+2{)}^{15}(t-4{)}^{6}(t+12{)}^{17}\, dt$, $x\in R$.
 If $p$ and $q$ are respectively the points of local minima and local maxima of $g$, then the value of $|p+q|$ is equal to \_\_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

The given equation is:

$f(x)=1-2x+{\int }_{x}^{0}{e}^{(x-t)}f(t)dt$

Since ${e}^{(x-t)}={e}^{x}\cdot {e}^{-t}$, we can move ${e}^{x}$ outside the integral because it is constant relative to $t$

$f(x)=1-2x+{e}^{x}{\int }_{x}^{0}{e}^{-t}f(t)dt$

Differentiate both sides with respect to $x$ using the Leibniz Rule and the product rule:

${f}^{'}(x)=-2+[{e}^{x}{\int }_{x}^{0}{e}^{-t}f(t)dt+{e}^{x}({e}^{-x}f(x))]$

Now, we use substitution from our original equation. Notice that ${e}^{x}{\int }_{x}^{0}{e}^{-t}f(t)dt=f(x)-(1-2x)$. Substituting this back into the derivative expression:

${f}^{'}(x)=-2+[f(x)-1+2x]+f(x)$

$\Rightarrow$ ${f}^{'}(x)=2f(x)+2x-3$

Combine the terms to get the linear differential equatio in the form of $\frac{df(x)}{dx}+P(x)f(x)=Q(x)$

$\frac{df(x)}{dx}-2f(x)=2x-3$

Solve for $f(x)$

Integrating Factor (IF): ${e}^{\int -2dx}={e}^{-2x}$

General Solution: $f(x)\cdot {e}^{-2x}=\int (2x-3){e}^{-2x}dx$

By integrating the right side (Integration by Parts), we get $f(x)=1-x+C{e}^{2x}$.

Using $f(0)=1$ (from the original equation), we find $C=0$.

Result: $f(x)=1-x$.

Find the Extrema of $g(x)$

$g(x)={\int }_{x}^{0}(f(t)+2{)}^{15}(t-4{)}^{6}(t+12{)}^{17}dt$

Differentiate both sides with respect to $x$ using the Leibniz Rule

${g}^{'}(x)=(f(x)+2{)}^{15}(x-4{)}^{6}(x+12{)}^{17}$

Substitute $f(x)=1-x$ :

$\begin{gathered} {g}^{'}(x)=(1-x+2{)}^{15}(x-4{)}^{6}(x+12{)}^{17} \ {g}^{'}(x)=(3-x{)}^{15}(x-4{)}^{6}(x+12{)}^{17} \end{gathered}$

Critical points: $x=3,4,-12$.

Plot on number line to find local maxima and minima.

Local Minima ( $p$ ): At $x=-12,{g}^{'}(x)$ changes sign from negative to positive. Thus, $p=-12$.

Local Maxima ( $q$ ): At $x=3,{g}^{'}(x)$ changes sign from positive to negative. Thus, $q=3$.

$x=4$ : No sign change occurs (even power 6), so it is not a local extremum.

Final Answer

$|p+q|=|-12+3|=|-9|=9$

The value of $|p+q|$ is 9 .

Solutions · 1