Hello Ann,
1.) We are given:
$$f(x)=\left(x+\frac{1}{x} \right)^2=x^2+2+x^{-2}$$
Although not necessary, I have expanded the function just to make differentiation a bit easier. So, differentiating, we find:
$$f'(x)=2x-2x^{-3}=\frac{2\left(x^4-1 \right)}{x^3}$$
Now, the tangent line at $x=a$ can be found since we have the point $\left(a,f(a) \right)$ and we have the slope $m=f'(a)$. Applying the point-slope formula, we get the equation of the tangent line as:
$$y-f(a)=f'(a)(x-a)$$
Arranging in slope-intercept form, we have:
$$y=f'(a)x+\left(f(a)-af'(a) \right)$$
Using $f(x)$ and $f'(x)$, we have:
$$f(a)=a^2+2+a^{-2}$$
$$f'(a)=\frac{2\left(a^4-1 \right)}{a^3}$$
and so, the tangent line, in terms of the parameter $a$, is:
$$y=\left(\frac{2\left(a^4-1 \right)}{a^3} \right)x+\left(a^2+2+a^{-2}-\frac{2\left(a^4-1 \right)}{a^2} \right)$$
Simplifying, we obtain:
$$y=\frac{2\left(a^4-1 \right)}{a^3}x+\frac{\left(1+a^2 \right)\left(3-a^2 \right)}{a^2}$$
Now, plugging in $$a=\frac{1}{2}$$, we get:
$$y=-15x+\frac{55}{4}$$
Here is a plot of the function and its tangent line:
View attachment 981
2.) We are given:
$$f(x)=e^{x}+e^{-x}$$
Differentiating, we find:
$$f'(x)=e^{x}-e^{-x}$$
Using the point $(a,f(a))$ and the slope $f'(a)$, we find the tangent line is:
$$y-f(a)=f'(a)(x-a)$$
Arranging in slope-intercept form, we have:
$$y=f'(a)x+\left(f(a)-af'(a) \right)$$
Using $f(x)$ and $f'(x)$, we have:
$$f(a)=e^{a}+e^{-a}$$
$$f'(x)=e^{a}-e^{-a}$$
and so, the tangent line, in terms of the parameter $a$, is:
$$y=\left(e^{a}-e^{-a} \right)x+\left(e^{a}+e^{-a}-a\left(e^{a}-e^{-a} \right) \right)$$
Simplifying, we obtain:
$$y=\left(e^{a}-e^{-a} \right)x+\left((1-a)e^{a}+(1+a)e^{-a} \right)$$
Now, plugging in $$a=0$$, we get:
$$y=2$$
Here is a plot of the function and its tangent line:
View attachment 982
