Answer:
a) f'(x) = 12x
b) -120
c) 36
d) y = 36x -55
Step-by-step explanation:
The derivative is the limit of the average slope on an interval as the interval size approaches zero. The equation for a tangent line is conveniently written in point-slope form when the tangent point and the derivative there are known.
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a)
The derivative is ...
[tex]\displaystyle f'(x)=\lim_{h\to 0}{\dfrac{f(x+h)-f(x)}{h}}=\lim_{h\to 0}\dfrac{(6(x+h)^2-1)-(6x^2-1)}{h}\\\\=\lim_{h\to 0}\dfrac{(6x^2+12xh+6h^2-1)-(6x^2-1)}{h}=\lim_{h\to 0}\dfrac{12xh+6h^2}{h}=\lim_{h\to 0}(12x+6h)\\\\\boxed{f'(x)=12x}[/tex]
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b)
f'(-10) = 12(-10) = -120
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c)
f'(3) = 12(3) = 36
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d)
The point of tangency is ...
(3, f(3)) = (3, 6·3² -1) = (3, 53)
The slope there was found in part (c) to be 36. So, the point-slope equation for the tangent line can be written ...
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
y -53 = 36(x -3) . . . . equation of the tangent line at (3, 53)
This equation can be rearranged to slope-intercept form:
y = 36x -55 . . . . . add 53 and simplify
