To find the derivative [tex]\( f^{\prime}(x) \)[/tex] of the function [tex]\( f(x) = \frac{3}{2}x^2 + 4x - 5 \)[/tex], we start with the definition of the derivative:
[tex]\[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \][/tex]
First, let's find [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x + h) = \frac{3}{2}(x + h)^2 + 4(x + h) - 5 \][/tex]
We need to expand and simplify [tex]\( f(x + h) \)[/tex]. Start by expanding [tex]\( (x + h)^2 \)[/tex]:
[tex]\[ (x + h)^2 = x^2 + 2xh + h^2 \][/tex]
Substitute back into the expression for [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x + h) = \frac{3}{2}(x^2 + 2xh + h^2) + 4(x + h) - 5 \][/tex]
Now distribute:
[tex]\[ f(x + h) = \frac{3}{2}x^2 + \frac{3}{2}(2xh) + \frac{3}{2}h^2 + 4x + 4h - 5 \][/tex]
Simplify further:
[tex]\[ f(x + h) = \frac{3}{2}x^2 + 3xh + \frac{3}{2}h^2 + 4x + 4h - 5 \][/tex]
Next, compute the difference [tex]\( f(x + h) - f(x) \)[/tex]:
[tex]\[ f(x + h) - f(x) = \left( \frac{3}{2}x^2 + 3xh + \frac{3}{2}h^2 + 4x + 4h - 5 \right) - \left( \frac{3}{2}x^2 + 4x - 5 \right) \][/tex]
Distribute and combine like terms:
[tex]\[ f(x + h) - f(x) = \frac{3}{2}x^2 + 3xh + \frac{3}{2}h^2 + 4x + 4h - 5 - \frac{3}{2}x^2 - 4x + 5 \][/tex]
[tex]\[ f(x + h) - f(x) = 3xh + \frac{3}{2}h^2 + 4h \][/tex]
Finally, divide by [tex]\( h \)[/tex] to obtain the difference quotient:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{3xh + \frac{3}{2}h^2 + 4h}{h} \][/tex]
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{h(3x + \frac{3}{2}h + 4)}{h} \][/tex]
Simplify by canceling [tex]\( h \)[/tex]:
[tex]\[ \frac{f(x + h) - f(x)}{h} = 3x + \frac{3}{2}h + 4 \][/tex]
This is the simplified form of the difference quotient.
In the next step, we will take the limit as [tex]\( h \)[/tex] approaches 0.
