Certainly! Let's go through the steps to solve the given problem involving the function [tex]\( f(x) = 2x^2 + 7x - 1 \)[/tex].
### Part (a): Simplifying the Difference Quotient
The difference quotient for the function [tex]\( f(x) \)[/tex] is given by:
[tex]\[ \frac{f(x+h) - f(x)}{h} \][/tex]
First, we find [tex]\( f(x+h) \)[/tex]:
[tex]\[ f(x+h) = 2(x+h)^2 + 7(x+h) - 1 \][/tex]
To expand [tex]\( f(x+h) \)[/tex], we perform the following steps:
[tex]\[ f(x+h) = 2(x+h)^2 + 7(x+h) - 1 \][/tex]
[tex]\[ f(x+h) = 2(x^2 + 2xh + h^2) + 7x + 7h - 1 \][/tex]
[tex]\[ f(x+h) = 2x^2 + 4xh + 2h^2 + 7x + 7h - 1 \][/tex]
Now, we subtract [tex]\( f(x) \)[/tex] from [tex]\( f(x+h) \)[/tex]:
[tex]\[ f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 7x + 7h - 1) - (2x^2 + 7x - 1) \][/tex]
Distributing the subtraction:
[tex]\[ f(x+h) - f(x) = 2x^2 + 4xh + 2h^2 + 7x + 7h - 1 - 2x^2 - 7x + 1 \][/tex]
Now, combine the like terms:
[tex]\[ f(x+h) - f(x) = (2x^2 - 2x^2) + 4xh + 2h^2 + (7x - 7x) + 7h - 1 + 1 \][/tex]
[tex]\[ f(x+h) - f(x) = 4xh + 2h^2 + 7h \][/tex]
Now, we divide by [tex]\( h \)[/tex] to get the difference quotient:
[tex]\[ \frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 7h}{h} \][/tex]
Simplify by canceling [tex]\( h \)[/tex] in the numerator and denominator:
[tex]\[ \frac{f(x+h) - f(x)}{h} = 4x + 2h + 7 \][/tex]
Thus, the simplified difference quotient is:
[tex]\[ \frac{f(x+h) - f(x)}{h} = 4x + 2h + 7 \][/tex]
### Part (b): Finding the Derivative
The derivative of the function, [tex]\( f'(x) \)[/tex], is the limit of the difference quotient as [tex]\( h \)[/tex] approaches zero:
[tex]\[ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \][/tex]
From our simplified difference quotient:
[tex]\[ \frac{f(x+h) - f(x)}{h} = 4x + 2h + 7 \][/tex]
Taking the limit as [tex]\( h \)[/tex] approaches zero:
[tex]\[ \lim_{h \rightarrow 0} (4x + 2h + 7) = 4x + 7 \][/tex]
Therefore, the derivative of the function is:
[tex]\[ f'(x) = 4x + 7 \][/tex]
