Question

For the function f(x) = x2 + 3, find the slope of the tangent line at x = 11.

Answer 1

Given:

The function is

$f(x)=x^2+3$

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,

$f(x)=x^2+3$

Differentiate with respect to x.

$f'(x)=2x+0$              $\left[\because \dfrac{d}{dx}x^n=nx^{n-1},\dfrac{d}{dx}C=0,\text{ where C is constant}\right]$

$f'(x)=2x$

Substitute x=11 in the above equation.

$f'(11)=2(11)$

$f'(11)=22$

Therefore, the slope of the tangent at x=11 is 22.