If
f(x) = 8x2 − x3,
find f '(2) and use it to find an equation of the tangent line to the curve
y = 8x2 − x3
at the point (2, 24).
So start by taking the derivative of f(x)
Then sub in f'(2) into the derivative you found : )
Awesome help oobleck : )
I am so happy we are on the same team : )
Yours in Mathematical fun, Ms Pi
Well, well, well, let's subtract seriousness from this equation and add some humor instead!
To find the derivative of f(x) = 8x^2 - x^3, we'll have to put on our math hats for a minute. Are you ready for some mathematical clowning around? Let's do this!
First, let's find the derivative. The derivative of f(x) with respect to x is denoted as f'(x) or dy/dx or clown-face-dance '/'. It represents the rate of change of the function.
f'(x) = 16x - 3x^2.
Now, let's plug in the value of x = 2 to find f'(2):
f'(2) = 16(2) - 3(2^2)
= 32 - 3(4)
= 32 - 12
= 20.
Now, with f'(2) = 20, we have the slope of the tangent line to the curve at the point (2, 24). Are you ready for the punchline? Here it comes!
Remember the point-slope equation formula? Y = mx + b, where m is the slope of the line and b is the y-intercept. Well, let's substitute in the values!
We have the slope m = 20 and the point (2, 24). Let's plug them in:
y - y1 = m(x - x1),
y - 24 = 20(x - 2),
y - 24 = 20x - 40,
y = 20x - 40 + 24,
y = 20x - 16.
Voila! We just found the equation of the tangent line to the curve y = 8x^2 - x^3 at the point (2, 24).
Remember, laughter is the best derivative in life!
To find the derivative, f'(x), of the function f(x) = 8x^2 - x^3, you need to apply the power rule and the constant multiple rule.
First, apply the power rule to each term in the function:
f'(x) = d/dx(8x^2) - d/dx(x^3)
The power rule states that the derivative of a term in the form x^n is d/dx(x^n) = nx^(n-1).
Using the power rule, you get:
f'(x) = 2 * 8x^(2-1) - 3x^(3-1)
= 16x - 3x^2
Now to find f'(2), substitute x = 2 into the derivative equation:
f'(2) = 16(2) - 3(2^2)
= 32 - 12
= 20
So, f'(2) = 20.
To find the equation of the tangent line to the curve y = 8x^2 - x^3 at the point (2, 24), we need both the slope and a point on the line.
The slope of the tangent line is equal to the derivative at that point, which we found to be f'(2) = 20.
Let's denote the equation of the tangent line as y = mx + b, where m is the slope and b is the y-intercept. We already have the value of the slope (m = 20).
To find the y-intercept (b), substitute the coordinates of the point (2, 24) into the equation:
24 = 20(2) + b
Simplifying, we get:
24 = 40 + b
Rearranging the equation to solve for b:
b = 24 - 40
b = -16
So, the equation of the tangent line to the curve y = 8x^2 - x^3 at the point (2, 24) is y = 20x - 16.
f' = 16x-3x^2
f'(2) = 32-24 = 8
now you have a point and a slope, so use the point-slope form of the line.