Definition of the derivative to calculate f'(x) if f(x) = square root 2x. Find the equation of the tangent & normal line to the graph of f at x=2.
Tangent line - y=2x-x
Normal line - y = 1/2x +3
I do not know whether these are the correct answers.
no.
y = √(2x) = (2x)^(1/2)
y' = (1/2)(2x)^(-1/2)(2) = 1/√(2x)
when x=2, y = 2, y' = 1/2
so the tangent line is
y = (1/2)x + b
which at (2,2) is 2 = (1/2)(2) + b
b = 1
so the tangent equation is y = (1/2)x + 1
the normal would have slope of -2
so at (2,2)
2 = -2(2) + b
b = 6
the normal has equation y = -2x + 6
Thank you . I think that I understand alittle better. It is oblivious that I do not understand this material.
I have one more that I really am lost at.
Question - find the points on the graph y=2x^3-3x^2-12x+20 at which the tangent is parallel to tthe x-axis. I do not even know where to begin. We were told we could use the shortcut for this one.!
Thanks again
If a tangent is parallel to the x-axis, wouldn't it have a slope of zero?
And isn't the slope represented by the first derivative?
so differentiate, set that equal to zero and solve for x
find the y's for those x's and simply say
y = (whatever the y values are)
let me know if you don't get y = 0 and y = 27
To find the derivative of a function, we can use the definition of the derivative. The derivative of a function f(x) represents the rate of change of f(x) at a specific point x.
In this case, we are given f(x) = √(2x). To find f'(x), we can use the definition of the derivative:
f'(x) = lim(h→0) [f(x+h) - f(x)] / h
Let's calculate step by step:
Step 1: Find f(x+h)
f(x+h) = √(2(x+h))
Step 2: Find f(x)
f(x) = √(2x)
Step 3: Find f(x+h) - f(x)
[f(x+h) - f(x)] = √(2(x+h)) - √(2x)
Step 4: Divide by h
[f(x+h) - f(x)] / h = [√(2(x+h)) - √(2x)] / h
Step 5: Take the limit as h approaches 0
lim(h→0) [√(2(x+h)) - √(2x)] / h
We can simplify the expression further using algebraic manipulations:
lim(h→0) [√(2(x+h)) - √(2x)] / h
Multiply by the conjugate to eliminate the radical:
lim(h→0) [√(2(x+h)) - √(2x)] / h * [√(2(x+h)) + √(2x)] / [√(2(x+h)) + √(2x)]
Simplifying gives:
lim(h→0) [2(x+h) - 2x] / [h (√(2(x+h)) + √(2x))]
Simplifying further:
lim(h→0) [2h] / [h (√(2(x+h)) + √(2x))]
Canceling out the common factor h:
lim(h→0) [2] / [√(2(x+h)) + √(2x)]
Now, substitute x = 2 into the equation to find f'(2):
f'(2) = [2] / [√(2(2+h)) + √(4)]
Simplifying the expression:
f'(2) = [2] / [√(4 + 2h) + 2]
This represents the value of the derivative of f(x) at x = 2. To find the equation of the tangent and normal lines at x = 2, we need both the slope and a point on the line.
Tangent Line:
We have the slope, which is f'(2). The point on the line can be found by evaluating f(2).
f(2) = √(2(2)) = √(4) = 2
So, the equation of the tangent line passing through (2, 2) is given by the point-slope form:
y = f'(2)(x - 2) + f(2)
= [2 / (√(4 + 2h) + 2)](x - 2) + 2
Normal Line:
The normal line is perpendicular to the tangent line and has a slope that is the negative reciprocal of the tangent line's slope.
The slope of the normal line is -1 / f'(2).
So, the equation of the normal line passing through (2, 2) is given by the point-slope form:
y = -1 / f'(2)(x - 2) + f(2)
= -1 / [2 / (√(4 + 2h) + 2)](x - 2) + 2
To obtain the specific equations of the tangent and normal lines, you'll need to calculate the value of f'(2) and further simplify the equations.