Answer
(a) $f'(2) = 10$
(b) $y = 10x-16$
(c) We can see the graph below.
Work Step by Step
(a) $f(x) = x^3-2x$
We can find $f'(x)$:
$\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\frac{[(x+h)^3-2(x+h)]-(x^3-2x)}{h}$
$=\lim\limits_{h \to 0}\frac{(x^3+3x^2h+3xh^2+h^3-2x-2h)-(x^3-2x)}{h}$
$=\lim\limits_{h \to 0}\frac{3x^2h+3xh^2+h^3-2h}{h}$
$=\lim\limits_{h \to 0}\frac{h(3x^2+3xh+h^2-2)}{h}$
$=\lim\limits_{h \to 0}(3x^2+3xh+h^2-2)$
$=3x^2-2$
Then:
$f'(2) = 3(2)^2-2 = 10$
(b) The tangent line at the point $(2,4)$ has a slope of $f'(2)$ which is $10$
We can find the equation of the tangent line:
$y - 4 = 10(x-2)$
$y-4 = 10x-20$
$y = 10x-16$
(c) We can see a graph of the curve, and the tangent line meeting the curve at the point $(2,4)$
