# CalculusConsider the equation below. f(x)= e^(3x) +e^-x (a) Find the intervals on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing....

Calculus

Consider the equation below.

f(x)= e^(3x) +e^-x

(a) Find the intervals on which *f* is increasing. (Enter your answer using interval notation.)

Find the interval on which *f* is decreasing. (Enter your answer using interval notation.)

(b) Find the local minimum value of *f*.

(c) Find the interval on which *f* is concave up. (Enter your answer using interval notation.)

### 1 Answer | Add Yours

f(x) = e^(3x)+e^(-x)

f'(x) = 3e^(3x) - e^(-x)

(a) Find the intervals on which *f* is increasing. (Enter your answer using interval notation.)

The function is increasing in the intervals where f'(x) (slope) is positive.

So f'(x)>0 when 3e^(3x) - e^(-x) > 0, 3e^(3x) > e^-x .

Dividing both sides by 3e^(-x) (which is always positive) we get

e^(4x) > 1/3, Taking the ln of both sides we get

4x > -ln(3)

x > (-ln(3))/4. So the function is increasing on the interval (-ln(3)/4, oo)

You could show that it is decreasing (f'(x)<0) on the interval (oo,-ln(3)/4)

Find the interval on which *f* is decreasing. (Enter your answer using interval notation.)

(b) Find the local minimum value of *f*.

The minimal can happen at the critical points f'(x) = 0, or the end points.

f'(x) = 0 (using the same steps as above) at x = -ln(3)/4

f''(x) = 9e^(3x) + e^-x and f''(-ln(3)/4)=9e^(-ln(3)/4)+e^(ln(3)/4) > 0 so

-ln(3)/4 is a local minima, and we can see that lim_(x->+-oo) f(x) = oo, so this is an absolute minima also.

(c) Find the interval on which *f* is concave up. (Enter your answer using interval notation.)

We computed f''(x) = 9e^(3x) + e^(-x). We can see that this is positive for all x, so this function is concave up (since f''(x)>0) on (-oo,oo).

The graph shows all of these things.