How do you sketch the graph that satisfies f'(x)0 when x does not equal 2, f(2)=1?

Question

How do you sketch the graph that satisfies f'(x)>0 when x does not equal 2, f(2)=1?

Emily Ammerman · Accepted Answer

I would apologize for being pedantic, but this is an educational website.

The question does not give any information about #f'(2)#. It is important to understand the use of language and logic in mathematics. Saying #f'(x) > 0# when #x != 2# DOES NOT entail that #f'(2) = 0# or even that #f'(2)# exists. #f'(2)# cannot be negative (because a function that is differentiable on an interval satisfies the intermediate value property), but it could be #0# or positive or it could fail to exist. So, (1) Any line with positive slope through #(2,1)# meets the condition for example #f(x)=3(x-2)+1# has #f'(x) = 3 > 0# for all #x != 2# and also for #x = 2# graph{y-1=3(x-2) [-1.907, 9.19, -2.45, 3.1]} Replace #3# with any positive number for another example. Any other curve with positive slope everywhere will also work, for example #f(x) - e^(x-2)# has positive slope everywhere and contains the point #(2,1)#. graph{e^(x-2) [-1.29, 8.574, -2.14, 2.793]} (2) We could also have a piecewise function with positive slope except at a discontinuity at #x=2#. #f(x) = {(5x+4,x != 2),(1,x = 2):}# This has #f'(x) = 5# for all #x != 2# and #f'(2) does not exist. Or #f(x) = (x-2)^(1/3)+1# which has #f'(x) = 1/2(x-2)^(-2/3) > 0# for #x !=2# and #f'(2)# does not exists (vertical tangent line). graph{(x-2)^(1/3)+1 [-2.68, 5.113, -0.86, 3.037]} (3) Or we could have a translation of an odd power function. #f(x) = (x-2)^3+1# or #f(x) = (x-2)^5+1# etc. These have #f'(x) >0 # for #x !=2# and #f'(2) =0# graph{(x-2)^(7/3)+1 [-1.216, 3.65, -0.128, 2.305]}

Elijah Abram · Answer

Since the derivative is greater than #0# on all #x# excluding #x = 2#, we know the function is increasing until it gets to #x= 2#, where it plateaus, and then it starts increasing again.

A perfect example of this would be the cubic function #f(x) = (x - 2)^3 + 1#, as pictured in the following graph.

Hopefully this helps!

Ezra Adams · Answer

undefined To sketch the graph of a function $ f(x) $ such that $ f'(x) > 0 $ when $ x \neq 2 $ and $ f(2) = 1 $, we follow these steps:

1. Determine the behavior of the function around $ x = 2 $:
   - Since $ f'(x) > 0 $ when $ x \neq 2 $, it implies that the function is increasing for all $ x $ values except at $ x = 2 $.
   - At $ x = 2 $, $ f(2) = 1 $, which means there is a point on the graph at $ (2, 1) $.

2. Sketch the graph:
   - Draw an increasing curve that passes through the point $ (2, 1) $.
   - The curve should approach but not touch the point $ (2, 1) $ from both sides.
   - Ensure that the curve is strictly increasing for $ x \neq 2 $.

3. Label the graph appropriately:
   - Label the point $ (2, 1) $ to indicate the specific value of the function at $ x = 2 $.
   - Optionally, include arrows or annotations to indicate the direction of the curve and the behavior of the function around $ x = 2 $.

4. Check the graph:
   - Ensure that the graph satisfies the conditions given: $ f'(x) > 0 $ for $ x \neq 2 $ and $ f(2) = 1 $.
   - Verify that the graph represents an increasing function for $ x \neq 2 $ and includes the specified point at $ (2, 1) $.

By following these steps, you can sketch the graph of the function that satisfies the given conditions.

How do you sketch the graph that satisfies f'(x)&gt;0 when x does not equal 2, f(2)=1?

How do you sketch the graph that satisfies f'(x)>0 when x does not equal 2, f(2)=1?