Find the equation of the tangent to the graph of f(x)=2x^4 (to the power of 4) that has slope 1.
Derivative is 8 x^3. This equals 1 at
x = 1/2. At this point the y-coordinate is f(1/2) = 1/8. The equation of the tangent line is thus:
1/8 + 1*(x-1/2) =
x - 3/8
Thank you for your assistance.
It clarified my thinking in regards to similar questions.
To find the equation of the tangent line to the graph of the function f(x) = 2x^4 that has a slope of 1, follow these steps:
1. Calculate the derivative of the function f(x). The derivative of 2x^4 with respect to x is 8x^3.
2. Set the derivative equal to 1, since the slope of the tangent line is given as 1:
8x^3 = 1
3. Solve for x by isolating it:
x^3 = 1/8
Taking the cube root of both sides, we get:
x = (1/8)^(1/3)
Simplifying further, we get:
x = 1/2
4. Find the y-coordinate at the point (1/2, f(1/2)). Substitute the value of x = 1/2 into the original function f(x):
f(1/2) = 2(1/2)^4
Simplifying further, we get:
f(1/2) = 1/8
Therefore, the coordinates of the point on the graph are (1/2, 1/8).
5. Use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) are the coordinates of the point on the graph and m is the slope of the tangent line.
Plugging in the values, we have:
y - (1/8) = 1(x - 1/2)
Simplifying further, we get:
y - 1/8 = x - 1/2
Rearranging the equation, we obtain the equation of the tangent line:
y = x - 1/2 + 1/8
Combining like terms, we finally get:
y = x - 3/8
Therefore, the equation of the tangent line to the graph of f(x) = 2x^4 with a slope of 1 is y = x - 3/8.
I'm glad I could assist you, and I hope this explanation helps you understand how to approach similar questions in the future. If you have any further questions, feel free to ask!