find the equation of tangent line to the graph of f(x)=xcosx(√2 x) at a=0
sorry the question is f(x)=xcos(√2 x) at a=0
well,
f'(x) = cos(√2 x) - √2 x sin(√2 x)
Not sure what a has to do with things, but if you meant x=0, then
f(0) = 0
f'(0) = 1
So, you want the line with slope 1, which passes through (0,0).
No too much of a challenge there, eh?
yep thanks
it will be y=x but how do we know it passes through (0,0)
Because y(0) = 0. It helps if you actually read what I write. Or maybe it has something to do with the mysterious "a".
okay I get it and a is as you said just to say x
To find the equation of the tangent line to the graph of the function f(x) = xcos(x√2) at a = 0, we can follow these steps:
Step 1: Find the derivative of the function f(x).
Step 2: Use the derivative to find the slope of the tangent line at x = a.
Step 3: Substitute the x-coordinate, a = 0, and the slope into the point-slope form of a line equation to obtain the final equation of the tangent line.
Let's start with Step 1:
Step 1: Find the derivative of f(x):
To find the derivative, we can use the product rule. The product rule states:
(uv)' = u'v + uv'
Applying this to the function f(x) = xcos(x√2):
f'(x) = [1 * cos(x√2)] + [x * (-sin(x√2)) * (√2)]
Simplifying:
f'(x) = cos(x√2) - x√2sin(x√2)
Step 2: Find the slope of the tangent line at x = a:
Evaluate the derivative at x = a (a = 0):
f'(0) = cos(0√2) - 0√2sin(0√2)
= cos(0) - 0
= 1
So, the slope of the tangent line at x = 0 is 1.
Step 3: Write the equation of the tangent line:
Now, we have the slope (m = 1) and a point (a, f(a)) = (0, f(0)). We can use the point-slope form of a line equation to write the equation of the tangent line:
y - y₁ = m(x - x₁)
Substituting in the values:
y - f(0) = 1(x - 0)
y - f(0) = x
Since a = 0, we have:
y - f(0) = x
y - (0cos(0√2)) = x
y = x
Thus, the equation of the tangent line to the graph of f(x) = xcos(x√2) at a = 0 is y = x.