Question

let g be the function defined by g(x) = ∫₀ˣ (-3/4 + t + cos(π/4 t² + t)) dt for 0 < x < 3. at what value of x does g attain a relative maximum?
a 1.025
b 1.361
c 1.616
d 2.310

Explanation:

Step1: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, if $g(x)=\int_{0}^{x}f(t)dt$, then $g^\prime(x) = f(x)$. Here, $f(t)=-\frac{3}{4}+t + \cos(\frac{\pi}{4}t^{2}+t)$.

Step2: Find critical points

Critical points of $g(x)$ occur where $g^\prime(x)=0$, so we need to solve $-\frac{3}{4}+t+\cos(\frac{\pi}{4}t^{2}+t)=0$.

Step3: Use a numerical method

Since this is a non - algebraic equation, we can use a graphing utility or a numerical method like Newton - Raphson method. Using a graphing calculator or software to graph $y =-\frac{3}{4}+t+\cos(\frac{\pi}{4}t^{2}+t)$ and find the $x$ - values (in the interval $0 < x<3$) where $y = 0$.
After using a graphing utility or numerical solver, we find that the root of $-\frac{3}{4}+t+\cos(\frac{\pi}{4}t^{2}+t)=0$ in the interval $(0,3)$ that corresponds to a relative maximum is approximately $x = 1.361$.
We can also check the sign of $g^\prime(x)$ around the critical points. If $g^\prime(x)$ changes sign from positive to negative at a critical point, then $g(x)$ has a relative maximum at that point.

Answer:

B. 1.361