Given f(x)=(3x-1)^2 Find the equation of the line tangent to the graph x=2, and find the value of x where the tangent line is horizontal
f(x) = (3x-1)^2 f ' (x) = 2(3x-1)^1 ... use the power rule now you need to multiply that with the derivative of the inside, so you need to derive 3x-1 to get 3, then you multiply 3 by that last expression on the right side to get f ' (x) = 3*2(3x - 1)^1 which simplifies to f ' (x) = 6(3x - 1) f ' (x) = 18x - 6
now plug in x = 2 into f ' (x) and tell me what you get
30
that's the slope of the tangent line at x = 2
so m = 2
I meant, m = 30
now we need to find the y coordinate of the point on f(x) when x = 2 f(x) = (3x-1)^2 f(2) = (3(2)-1)^2 ... plug in x = 2 f(2) = (6-1)^2 f(2) = 5^2 f(2) =25
so the point (2,25) is on the function f(x)
we have enough to find the equation of the tangent line at x = 2 this is because we have a slope (m = 30) and a point the line goes through (2,25)
y = mx + b y = 30x + b 25 = 30(2) + b solve for b
tell me what you get
-35
so your tangent line is y = 30x - 35
tangent line at x = 2
that answers the first part of the question
f ' (x) = 18x - 6 represents the slope of the tangent line at ANY point on f(x)
if the tangent line is horizontal, then the slope is 0
so when we want to find what values of x make horizontal tangent lines, we just plug in f ' (x) = 0 and solve for x f ' (x) = 18x - 6 0 = 18x - 6 ... ... x = ??
3
close
.3
but it's a bit off
repeating
good or x = 1/3
at x = 1/3, the tangent line is horizontal
finding horizontal tangents is useful for maximizing or minimizing a function so you can apply it to maximizing profit or minimizing costs
well...that seems very easy
yeah it's not too bad once you get to know it
that seemed like algebra easy...like I was WAY overthinking it!
well you're definitely using calculus ideas and rules, but in a way, yes it's like algebra
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