So, I was studying for a test and came across the following question:
"Let $$g(x)$$ be the function by $g(x)= \int_0^x (\frac{-3}{4}+t+cos(\frac{pi}{4}*t^2+t))$ for $0<x<4$. At what value of $x$ does $g$ attain a relative maximum?"
MathJax doesn't seem to wanna work on my computer, so here's a screenshot of the prompt as well:
I'm a bit confused on how I'd go about tackling something like this, would it involve converting the integral into terms of $x$? Any help would be appreciated!
The fundamental theorem of calculus "says" that $$g(x)=\int_0^x f(t) \, dt \implies g'(x)=f(x)$$ So, you first need to find the zero('s) of $$g'(x)=-\frac{3}{4}+x+\cos\left(\frac{\pi}{4}x^2+x\right)$$ This is a transcendental equation which, by definition, will not show analytical solutions. As a consequence, some numerical methods will be required and then a reasonable starting guess.
If you plot the function, there is a root which is very close to $x=1.0$ and another close to $x=1.6$
Now, you are ready for Newton method which will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.000000 \\ 1 & 1.024501 \\ 2 & 1.024508 \end{array} \right)$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.600000 \\ 1 & 1.616232 \\ 2 & 1.615677 \end{array} \right)$$ One of them will be a maximum and the other will be a minimum.