Find the equation of the tangent to the graph y=f'(x) at x=3 where f(x)=x^4
So I did :
f'(x)=4x^3
y-81=4x^3(x-3)
y=4x^4-12x+81
?? Am i even remotely correct?
plug in 4 for x
f'(x)=4(4)^3 =?
I mean plug in 3 for x
f'(x)=4(3)^3 =?
108
y-81=108(x-3)
and solve for y.
y=108x-243
Yes, you are on the right track! To find the equation of the tangent to the graph of y = f'(x) at x = 3, you need to follow these steps:
1. Find the derivative of the given function f(x) = x^4. The derivative of x^4 can be found using the power rule of differentiation. Differentiating each term of the function gives: f'(x) = 4x^3.
2. Next, substitute the given x-coordinate, x = 3, into the derivative equation to find the slope of the tangent line. Plug 3 into the derivative: f'(3) = 4(3)^3 = 108.
3. Now that you have the slope of the tangent line, you also need a specific point on the line to determine its equation. Since the tangent line is touching the graph of y = f'(x) at x = 3, substitute x = 3 into the original function f'(x) to find the corresponding y-coordinate: f'(3) = 4(3)^4 = 324.
4. Now that you have the slope (m = 108) and a point (3, 324) on the tangent line, you can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by: y - y1 = m(x - x1).
Using the values from step 3, the equation of the tangent line becomes: y - 324 = 108(x - 3).
Now simplify and rewrite the equation in standard form: y - 324 = 108x - 324. Rearranging the terms gives: y = 108x.
Thus, the equation of the tangent line to the graph of y = f'(x) at x = 3 is y = 108x.