## Calculus: Early Transcendentals 8th Edition

$f'(x)=2-\frac{15}{4}x^{-1/4}$ $f''(x)=\frac{15}{16}x^{-5/4}$ These functions are reasonable since f' is positive when f is increasing and negative when f is decreasing. The same goes for f'' and f' - since f' is always increasing, f'' is always positive. The slopes of the derivatives are closer to zero wherever the graphs of the original functions appear almost flat.
$f(x)=2x-5x^{3/4}$ Use the Power Rule: $f(x) = x^{n},$ then $f'(x) = nx^{n-1}$ $f'(x)=2(1)-5(3/4)x^{-1/4}$ $f'(x)=2-\frac{15}{4}x^{-1/4}$ $f''(x)=0-(\frac{15}{4})(-\frac{1}{4})x^{-5/4}$ $f''(x)=\frac{15}{16}x^{-5/4}$