# Question 80732

Sep 15, 2016

Just stay calm. f'(x) = 3x - 1
Explanation

$f \left(x + h\right) = 1.5 {\left(x + h\right)}^{2} - \left(x + h\right) + 3.7 = 1.5 {x}^{2} + 3 x h + 1.5 {h}^{2} - x - h + 3.7$
$f \left(x + h\right) = f \left(x\right) + 3 x h + 1.5 {h}^{2} - h$
$\frac{f \left(x + h\right) - f \left(x\right)}{h} = 3 x - 1 + 1.5 h$

Now take the limit h to 0. The last term drops and you get
$f ' \left(x\right) = \lim h \to 0 \frac{f \left(x + h\right) - f \left(x\right)}{h} = 3 x - 1$
Explanation:
f(x+h)=1.5(x+h)^2−(x+h)+3.7=1.5x^2+3xh+1.5h^2−x−h+3.7#

$\frac{f \left(x + h\right) - f \left(x\right)}{h} = 3 x - 1 + 1.5 h$
Taking the limit gives the above result.