Suppose that g(x) is a function that is defined and differentiable for all real numbers x and that g(x) has the following properties:
(i) g(0)=5 and g'(0)=-5/6 .
(ii) g(7)=6 and g'(7)=6 .
(iii) g(x) is concave up for x<7 and concave down for x>7.
(iv) g(x)>_-15 for all x.
Use these properties to answer the following questions.
(a) How many zeros does g have?
(b) How many zeros does g' have?
sorry - once is enough. But
(iii) says that g''(7) = 0
so start with g'' = a(x-7) with a < 0
(iv) says that there will be either 1 or 3 zeroes (why?)
Now you try something.
To answer these questions, we can make use of the properties given for the function g(x) and its derivative g'(x). Let's break down each question:
(a) How many zeros does g have?
Since g(x) is defined for all real numbers and g(x) is a continuous function, we can use the Intermediate Value Theorem (IVT) to determine the number of zeros of g(x). The IVT states that if a function is continuous on a closed interval [a, b] and takes on values f(a) and f(b) with opposite signs, then there exists at least one value c in the interval (a, b) such that f(c) = 0.
From the given property (iv), we know that g(x) > -15 for all x. This means that g(x) is always positive and never crosses the x-axis, thereby having zero solutions. Therefore, g(x) has 0 zeros.
(b) How many zeros does g' have?
To determine the number of zeros of g'(x), we need to identify where g'(x) changes sign. Since g'(x) is the derivative of g(x), its sign indicates the slope and behavior of g(x). The changing sign of g'(x) implies that g(x) changes from increasing to decreasing or vice versa.
From the given properties (i) and (ii), we know the values of g'(0) and g'(7). It is given that g'(0) = -5/6 and g'(7) = 6. As g(x) is concave up for x < 7 and concave down for x > 7, this indicates a change in slope at x = 7.
Since g'(0) is negative and g'(7) is positive, g'(x) changes sign from negative to positive at x = 7. Therefore, g'(x) has one zero at x = 7.
In summary:
(a) g(x) has 0 zeros.
(b) g'(x) has 1 zero at x = 7.