If
f(x) = 9x2 − x3,
find f'(2) and use it to find an equation of the tangent line to the curve
y = 9x2 − x3
at the point (2, 28).
To find f'(2), we first need to differentiate the function f(x) with respect to x.
f(x) = 9x^2 - x^3
To differentiate, we apply the power rule of differentiation as follows:
f'(x) = d/dx(9x^2) - d/dx(x^3)
Using the power rule, we can find the derivatives of the terms individually:
d/dx(9x^2) = 2 * 9 * x^(2-1) = 18x
d/dx(x^3) = 3 * x^(3-1) = 3x^2
Now we can find f'(x):
f'(x) = 18x - 3x^2
To find f'(2), substitute x = 2 into the derivative:
f'(2) = 18(2) - 3(2^2)
= 36 - 3(4)
= 36 - 12
= 24
Next, we need to find the equation of the tangent line to the curve y = 9x^2 - x^3 at the point (2, 28).
The equation of a straight line can be written as y = mx + b, where m is the slope and b is the y-intercept.
We already have the slope of the tangent line, which is f'(2) = 24. To find the y-intercept, we substitute the coordinates of the point (2, 28) into the equation y = mx + b:
28 = 24(2) + b
Solving for b, we have:
28 = 48 + b
b = 28 - 48
b = -20
Therefore, the equation of the tangent line to the curve y = 9x^2 - x^3 at the point (2, 28) is:
y = 24x - 20.
To find the derivative of a function and use it to find the equation of the tangent line to the curve, follow these steps:
Step 1: Find the derivative of the function f(x) = 9x^2 - x^3. This will give you the slope of the tangent line.
To find the derivative, you can use the power rule for derivatives. The power rule states that if you have a term of the form x^n, the derivative is given by nx^(n-1).
For f(x) = 9x^2 - x^3, apply the power rule to each term:
f'(x) = 2(9x^(2-1)) - 3x^(3-1)
= 18x - 3x^2
Step 2: Evaluate the derivative at x = 2 to find f'(2). This will give you the slope of the tangent line at the point (2, 28).
Substitute x = 2 into the derivative expression:
f'(2) = 18(2) - 3(2^2)
= 36 - 12
= 24
So, f'(2) = 24.
Step 3: Use the point-slope form of the equation of a line to find the equation of the tangent line at the point (2, 28). The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Let's substitute the point (2, 28) and the slope f'(2) = 24 into the point-slope form:
y - 28 = 24(x - 2)
Simplifying the equation:
y - 28 = 24x - 48
Rearranging the equation to slope-intercept form (y = mx + b):
y = 24x - 20
So, the equation of the tangent line to the curve y = 9x^2 - x^3 at the point (2, 28) is y = 24x - 20.