Did you try anything ?
1. f(x) = (2x-1)^5 (x+1)
What is the equation of the tangent line at x=1?
a. y=21x+2
b. y=21x-19
c. y=11x-9
d. y=10x+2
e. y=10x-8
2. f ' (x) = 2x+1 f(1) = 4
Approximate f(1.2) using line tangent to the graph at x=1
a. 0.6
b. 3.4
c. 4.2
d. 4.6
e. 4.64
I appreciate any help.
Thanks.
The tangent line to y= f(x) at x= a is y= f'(a)(x- a)+ f(a).
What is the derivative of (2x-1)^5(x+1) at x= 1?
You can approximate f(1.2), with f'(x)= 2x+ 1 and f(1)= 4, by taking the tangent line, at x= 1, evaluated at x= 1.2. What is f'(1) ?