Find an equation of the tangent line at x = 0 for the given function.
f(x) = 5x2 + cos(x)
I really need help on this one.
Differentiate f(x) with respect to x.
Evaluate f'(x=0) for the slope.
Find the equation of a straight line with slope f'(0) that passes through (0,f(0)).
f(x) = 5x² + cos(x)
f'(x)= 10x - sin(x)
f'(0) = 10(0) - sin(0) = 0-0=0
f(0)=5(0)²+cos(0)=0+1=1
We now look for a straight line that passes through (x0,y0)=(0,f(0))=(0,1) with a slope of f'(0), i.e.
(y-y0)=m(x-x0)
y-1 = 0(0-0)
y=1 ( a horizontal line passing through (0,1) )
o, ok thank you!
I now understand.
To find the equation of the tangent line at x = 0 for the function f(x) = 5x^2 + cos(x), we need to follow these steps:
Step 1: Find the derivative of the function f(x).
The derivative of f(x) will give us the slope of the tangent line at any point x on the graph of f(x).
Let's start by finding the derivative d/dx of each term individually:
- The derivative of the first term 5x^2 is 10x.
- The derivative of the second term cos(x) is -sin(x).
Now, taking the sum of these derivatives, we get the derivative of the function f(x):
f'(x) = 10x - sin(x)
Step 2: Evaluate the derivative at x = 0 to find the slope of the tangent line.
When we substitute x = 0 into the derivative expression, we get:
f'(0) = 10(0) - sin(0)
= -sin(0)
= 0
The slope of the tangent line at x = 0 is 0.
Step 3: Determine the y-coordinate of the point of tangency.
Substitute x = 0 into the original function f(x) to find the y-coordinate of the point of tangency.
f(0) = 5(0)^2 + cos(0)
= 0 + 1
= 1
The y-coordinate of the point of tangency is 1.
Step 4: Write the equation of the tangent line in point-slope form.
Using the slope (m = 0) and the point of tangency (0, 1), we can write the equation of the tangent line in point-slope form:
y - y1 = m(x - x1)
Substituting the known values, we get:
y - 1 = 0(x - 0)
y - 1 = 0
y = 1
Thus, the equation of the tangent line at x = 0 for the function f(x) = 5x^2 + cos(x) is y = 1, a horizontal line passing through the point (0, 1).