Given:
The function is
[tex]f(x)=x^2+3[/tex]
To find:
The slope of the tangent line at x = 11.
Solution:
The slope of the tangent is the value of f'(x) at x=11.
We have,
[tex]f(x)=x^2+3[/tex]
Differentiate with respect to x.
[tex]f'(x)=2x+0[/tex] [tex]\left[\because \dfrac{d}{dx}x^n=nx^{n-1},\dfrac{d}{dx}C=0,\text{ where C is constant}\right][/tex]
[tex]f'(x)=2x[/tex]
Substitute x=11 in the above equation.
[tex]f'(11)=2(11)[/tex]
[tex]f'(11)=22[/tex]
Therefore, the slope of the tangent at x=11 is 22.
