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Sagot :
The given function is
f(x) = 6x^4 + 6
The formula for the limit is shown below
[tex]\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{ }\frac{6(x+h)^4+6-(6x^4+6)}{h} \\ f^{\prime}(x)\text{ = }\lim _{h\to0}\text{ }\frac{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{ }\frac{h(6h^3+24h^2x+36hx^2+24x^3)}{h} \\ h\text{ cancels out} \\ \end{gathered}[/tex]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)
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