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) )
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.
2 answers
o, ok thank you!
I now understand.
I now understand.