Problem:$\displaystyle \displaystyle \lim_{ \Delta x \to 0} \frac{[5-3(1-\Delta x)]- 2}{ \Delta x}$
I got
$\displaystyle f(x)= 2+3x \quad
f'(x)= 3 \quad
c=0 \quad$
Is this correct?
I assume you are trying to find the derivative of f(x) = 2+3x.
From the definition of the derivative:
$\displaystyle
f'(x) = \displaystyle\lim_{\Delta x \to 0} \frac {f(x+\Delta x) - f(x)} {\Delta x} = \displaystyle\lim_{\Delta x \to 0} \frac {(2+3(x+\Delta x)) - (2 + 3x)} {\Delta x} = \displaystyle\lim_{\Delta x \to 0} \frac {3 \Delta x} {\Delta x} = 3
$
Not sure what you mean by "c" or why you posted the limit that you did. It looks like it might be an attempt at an expression for the limit of the function 5+3x evaluated at x = -1?
OK. The "alternate form" of the derivative is:
$\displaystyle
f'(x) = \displaystyle \lim_{x \to c} \frac {f(x)-f(c)} {x-c}
$
Here it looks like they have changed it slightly, using $\displaystyle c-x = \Delta x$. It appears that the function may be $\displaystyle f(x) = 5-3x$, with $\displaystyle c = 1$. This makes $\displaystyle f(c) = 2$, and the alternate form is then:
$\displaystyle
f'(1) = \displaystyle \lim_{x \to 1} \frac {5-3x - 2} {x-1}
$
Now substituting $\displaystyle x = c- \Delta x = 1 - \Delta x$:
$\displaystyle
\displaystyle \lim_{x \to 1} \frac {5-3(1-\Delta x)-2} {\Delta x}
$
Which is what you were given. To evaluate the derivative:
$\displaystyle
f'(x) = \displaystyle \lim_{x \to c} \frac {(5-3x) - (5-3c)} {x-c} = \displaystyle \lim_{x \to c} \frac {-3x+3c} {x-c} = \displaystyle \lim_{x \to c} \frac {-3(x-c)} {x-c} = -3
$
So: $\displaystyle f(x) = 5-3x$, $\displaystyle c=1$, $\displaystyle f'(x) = -3$