# Question 6

## The use of a graphing calculator is **NOT** allowed.

The figure shows ƒ', the derivative of a function ƒ.

The domain of the function ƒ is the interval [0, 20].

(a) For what values of *x*, 0 < *x* < 20, does ƒ have a relative maximum? Justify your answer.

(b) For what values of *x* is the graph of ƒ concave down? Justify your answer.

(c) If ƒ(0) = 10, sketch a graph of the function ƒ on axes similar to the ones provided below. List the coordinates of all critical points and inflection points.

**Solution**

The relative maximum depends on the where the graph of the derivative of **f** equals zero. Due to the fact that the function **f**, exists from 0 to 20 including both endpoints. Between 0 and 20, the maximum will be visible on **f'** because it will be above the x axis before it **crosses** the x axis and exists below the x axis.

There are three x coordinates that need to be considered for this problem. The obvious x=10, and the endpoints, x=0 and x=20. No one knows if x=10 is greater than x=0, but if you take the integral from 0 to 10, you will find that out.

Σ(this takes the place of the integral sign)

Σ(0,10) Can be found by finding the area of a triangle.

A = b(h)/2

A = 10(10)/2

A = 50

Because A between 0 and 10 is positive,when x=10 the value of **f** is **greater** than when x=0.

Now that x=0 has been eliminated as a choice for maximum, the other contender needs to be compared to x=10.

Take the integral from 10 to 20, and if it is positive, then x=20 is the max, if negative, x=10 is the max.

Σ(10,15) is a triangle.

A = b(h)/2

A = 5(-5)/2

A = -12.5

Σ(15,20) is also a triangle.

A = b(h)/2

A = 5(10)/2

A = 25

Σ(10,20)(f'(x))= 25-12.5

Σ(10,20)(f'(x))= 12.5

Because the integral from 10 to 20 is positive, the relative maximum occurs when x=20.

**Answer to Part a) The relative maximum occurs when x=20.**

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