An equation of the line tangent to the graph of f(x) = (4x2 - 8x + 3)4 at the point where x = 1 is:
y=1
thx anon is right welp that's what apex said
Well, if I were to describe the equation of the line tangent to that graph, I would say it's "straightforward"! But seriously, let me calculate that for you.
To find the equation of a tangent line, we need the slope of the tangent line and the coordinates of a point on that line. So, let's find the values first.
To find the slope, we can differentiate f(x) = (4x^2 - 8x + 3)^4 with respect to x. After a few calculations, we get f'(x) = 16(4x^2 - 8x + 3)^3 * (8x - 8). Now, let's find the slope at x = 1.
Plug-in x = 1 into f'(x): f'(1) = 16(4(1)^2 - 8(1) + 3)^3 * (8(1) - 8).
After some more calculations, we find f'(1) = 0. So, the slope of the tangent line is 0.
Now, since the line is tangent at x = 1, we know that the point on the line is (1, f(1)). Let's find the value of f(1).
Plug-in x = 1 into f(x): f(1) = (4(1)^2 - 8(1) + 3)^4.
This simplifies to f(1) = (-1)^4 which is 1. So, the coordinates of the point on the tangent line are (1, 1).
Putting it all together, the equation of the tangent line can be written in the form y = mx + b, where m is the slope and b is the y-intercept. Since the slope is 0, we can represent the equation of the line as y = 1.
So, the equation of the line tangent to the graph of f(x) = (4x^2 - 8x + 3)^4 at the point where x = 1 is y = 1.
Isn't math punny?
To find the equation of the line tangent to the graph of the function f(x) at a specific point, you need to follow these steps:
Step 1: Find the derivative of the function f(x).
Step 2: Substitute the x-coordinate of the point of tangency into the derivative to find the slope of the tangent line.
Step 3: Use the point-slope form of the equation of a line, y - y₁ = m(x - x₁), where (x₁, y₁) represents the given point of tangency and m represents the slope of the tangent line.
Let's go through each step:
Step 1: Find the derivative of f(x).
The function f(x) = (4x^2 - 8x + 3)^4 can be simplified by applying the chain rule to the outer exponent 4 and the inner function (4x^2 - 8x + 3).
The derivative of f(x) can be found by multiplying the derivative of the outer exponent (4u^3) by the derivative of the inner function (8x - 8), where u = 4x^2 - 8x + 3.
So, f'(x) = 4(4x^2 - 8x + 3)^3 * (8x - 8).
Step 2: Substitute x = 1 into f'(x) to find the slope of the tangent line.
Substituting x = 1 into f'(x), we have:
f'(1) = 4(4(1)^2 - 8(1) + 3)^3 * (8(1) - 8)
= 4(4 - 8 + 3)^3 * (8 - 8)
= 4(-1)^3 * 0
= 0.
The slope of the tangent line is 0.
Step 3: Use the point-slope form of the equation of a line.
We have the point of tangency as (1, f(1)), where f(1) is the value of the function f(x) at x = 1.
Substituting the slope (m = 0) and the point (x₁ = 1, y₁ = f(1)) into the point-slope form, we have:
y - f(1) = 0(x - 1).
Since the slope is 0, the equation simplifies to:
y - f(1) = 0.
Therefore, the equation of the line tangent to the graph of f(x) = (4x^2 - 8x + 3)^4 at the point where x = 1 is simply:
y = f(1).
To find the value of f(1), substitute x = 1 into the original function:
f(1) = (4(1)^2 - 8(1) + 3)^4
= (4 - 8 + 3)^4
= (-1)^4
= 1.
So, the equation of the line tangent to the graph of f(x) = (4x^2 - 8x + 3)^4 at the point where x = 1 is:
y = 1.
dy/dx = 4(8 x -8) = 32 x -32
when x = 1, slope = 0 (horizontal line)
y = 4(4-8+3) = 4(-1) = -4
so the equation is y = -4