Answer to Question #273982 in Calculus for Josa Kyme Palma

Question

Find all relative extrema. Use the second derivative test where applicable.

(a) f(x)=√x2+1

(b) f(x)=cos x -x

Answer 1

(a) $f(x)=\sqrt{x^2+1}=(x^2+1)^{1\over 2}$

$f'(x)={1\over 2}(x^2+1)^{-{1\over 2}}.2x={x\over \sqrt {x^2+1}}$

Equate $f'(x)$ to to get the value of $x$

$f'(x)={1\over 2}(x^2+1)^{-{1\over 2}}.2x={x\over \sqrt {x^2+1}}=0 \implies x=0$

$f'(x)\lt 0$ for $x\lt 0$ and $f'(x)\gt0$ for $x\gt 0$ $\implies$ at $x=0$ there is a minimum.

Because $f(x)$ was decreasing till it reaches a minimum at $x=0$ and then started increasing ,because the derivative is positive after $x=0$

(b) $f(x)=cos( x) -x$

$f'(x)=-sin(x)-1$

Equate $f'(x)$ to to get the value of $x$

$f'(x)=-sin(x)-1 =0 \implies x={3\pi \over 2}$

$f''(x)=-cos(x)$

$f''({3\pi \over 2})=-cos({3\pi \over 2})=0 \implies$ we have an inflection point at $x={3\pi \over 2}$

so,No extrema anywhere