Question

f(x) = 6x^4 + 6Use the limit process to find the slope of the line tangent to the graph of f at x = 2. Slope at x= 2:__Find an equation of the line tangent to the graph of f at x = 2:__

Answer 1

The given function is

f(x) = 6x^4 + 6

The formula for the limit is shown below

`\begin{gathered} f^(\prime)(x)\text{ = }\lim _(h\to0)\text{ }\frac{f(x\text{ + h) - f(x)}}{h} \\ \text{Substituting x = x + h into the function, we have} \\ f^(\prime)(x)\text{ = }\lim _(h\to0)\text{ }(6(x+h)^4+6-(6x^4+6))/(h) \\ f^(\prime)(x)\text{ = }\lim _(h\to0)\text{ }(6(h^4+4h^3x+6h^2x^2+4hx^3+x^4)+6-6x^4-6)/(h) \\ f^(\prime)(x)\text{ = }\lim _(h\to0)\text{ }\frac{6h^4+24h^3x+36h^2x^2+24hx^3+6x^{4\text{ }}-6x^4\text{ + 6 - 6}}{h} \\ f^(\prime)(x)\text{ = }\lim _(h\to0)\text{ }(h(6h^3+24h^2x+36hx^2+24x^3))/(h) \\ h\text{ cancels out} \\ \end{gathered}`

Evaluating the limit at h = 0, we would substitute h = 0 into 6h^3 + 24h^2x + 36hx^2 + 24x^3

It becomes

6(0)^3 + 24(0)^2x + 36(0)x^2 + 24x^3

The derivative is 24x^3

f'(x) = 24x^3

This is the slope of the tangent line is at x = 2

By substituting x = 2 into f'(x) = 24x^3, it becomes

f'(2) = 24(2)^3 = 192

To find the y coordinate of the point, we would substitute x = 2 into

f(x) = 6x^4 + 6

y = 6(2)^4 + 6 = 102

Thus, the x and y coordinates are (2, 102) and the slope is 192

The equation of the line in the point slope form is

y - y1 = m(x - x1)

Thus, the equation of the tangent is

y - 102 = 192(x - 2)