Find the values for which f(x) = x^10-10x and if 0 (less than or equal) to x (which is less than or equal to) 2.
a) f(x)has a local max or local min. Indicate which ones are max and which are mins.
b) f(x) has a global max or global min
could you kind of give me a hint on how to start and solve this question? i'm not really sure how to start working on it.
0 </= x </= 2 is the domain?
y = x^10 - 10 x
if you graph this it will start at 0, drop below axis, come back up to 0 when x^9 = 10 (about x = 1.3) and take off like mad up to the right
so part a
dy/dx = 10 x^9 - 10
zero when x^9 = 1
or when x = 1
that is in our domain and we already know it is a min but to prove it find y"
y" = 90 x^8 where x = 1
so y" is positive so this is a minimum at x = 1
there
y = 1^10 - 10 = -9
so
(1, -9) is local min
local max will be where x = 2 if y>0 there
y = 2^10 - 20
y = 1024 - 20 = 1004
so
(2, 1004) is the maximum in the domain
outside our domain x^10 is even so goes to oo for negative x and the -10x term is positive for negative x so it never drops below zero to the left of the origin and has no minima before heading off to +infinity for large -x
thank you so much, that makes so much more sense now!!!
To solve this question, we need to find the local maxima or minima and global maxima or minima of the given function f(x) = x^10 - 10x over the interval [0,2].
To begin, let's first find the critical points of the function. Critical points occur where the derivative of the function is either zero or undefined.
Step 1: Find the derivative of f(x)
f'(x) = d/dx (x^10 - 10x)
Using the power rule of differentiation, f'(x) = 10x^9 - 10
Step 2: Find the critical points
To find the critical points, we set the derivative equal to zero and solve for x:
10x^9 - 10 = 0
Dividing both sides by 10, we get:
x^9 - 1 = 0
Step 3: Solve for x
x^9 = 1
To solve for x, we take the ninth root of both sides:
x = 1
So, we have found one critical point at x = 1.
Step 4: Check the endpoints of the interval
Since we are interested in the interval [0,2], we should also check the function values at the endpoints.
f(0) = 0^10 - 10*0 = 0
f(2) = 2^10 - 10*2 = 1024 - 20 = 1004
Now, let's analyze each part of the question:
a) Local maxima or minima:
To determine whether the critical point at x = 1 is a local maximum or minimum, we can perform a second derivative test. However, since the second derivative of the given function is always positive, it means the function is always concave up.
Since the function is concave up, the critical point is at x = 1 is a local minimum.
b) Global maxima or minima:
To determine whether the local minimum at x = 1 is also the global minimum, we need to compare the function values at the endpoints.
Since f(0) = 0 and f(2) = 1004, we observe that f(2) is greater than f(0). Therefore, the global minimum occurs at x = 0 on the given interval.
In summary,
a) f(x) has a local minimum at x = 1.
b) f(x) has a global minimum at x = 0.
I hope this explanation helps you understand the process of solving this question!