How do you find the local max and min for #f(x) = 7x + 9x^(-1)#?

Question

Paisley Johnson · Accepted Answer

To find the local extrema we find the points where the first derivative is null and study the sign of the second derivative

#f(x) =7x+9x^(-1)# #f'(x) =7 -9x^(-2)# #f''(x) =18x^(-3)# Find the values of #x# where #f'(x)=0# #7-9/x^2 = 0# #7x^2-9 = 0# #x=+-3/sqrt(7)# In both points #f''(x) !=0# so these are local extrema, namely: 1) around #x=-3/sqrt(7)# the second derivative is negative and then the point is a local maximum. 2) around #x=+3/sqrt(7)# the second derivative is positive and then the point is a local minimum. graph{7x+9/x [-47.3, 47.2, -73.6, 73.6]}

Ezra Adams · Answer

undefined To find the local maximum and minimum for the function $ f(x) = 7x + 9x^{-1} $, you first need to find its derivative and then solve for critical points.

1. Find the derivative of $ f(x) $:
$$ f'(x) = 7 - 9x^{-2} $$

2. Set the derivative equal to zero to find critical points:
$$ 7 - 9x^{-2} = 0 $$
$$ 9x^{-2} = 7 $$
$$ x^{-2} = \frac{7}{9} $$
$$ x^2 = \frac{9}{7} $$
$$ x = \pm \sqrt{\frac{9}{7}} $$

3. To determine if these critical points correspond to local maximum or minimum, you can use the second derivative test or evaluate the sign of the derivative around these points.

4. Evaluate the second derivative:
$$ f''(x) = 18x^{-3} $$

5. Plug the critical points into the second derivative:
$$ f''\left(\sqrt{\frac{9}{7}}\right) = 18\left(\frac{9}{7}\right)^{-\frac{3}{2}} $$
$$ f''\left(-\sqrt{\frac{9}{7}}\right) = 18\left(-\frac{9}{7}\right)^{-\frac{3}{2}} $$

6. Determine the concavity of the function at these critical points. If $ f''(x) > 0 $, it's concave up, indicating a local minimum. If $ f''(x) < 0 $, it's concave down, indicating a local maximum.

7. Once you determine the concavity at these critical points, you can conclude whether they correspond to local maximum or minimum values for $ f(x) $.