The graph y= f(x) passes through the points (1, 5) and (3, 7). The tangent line to y= f(x) at (3, 7) has the equation: y= -2x + 13. The graph y= f(x) passes through the points (1, 5) and (3, 7)....

The graph y= f(x) passes through the points (1, 5) and (3, 7). The tangent line to y= f(x) at (3, 7) has the equation: y= -2x + 13.

The graph y= f(x) passes through the points (1, 5) and (3, 7). The tangent line to y= f(x) at (3, 7) has the equation: y= -2x + 13. make a possible graph of f and the tangent line. What is the average rate of change of  f(x) on the interval 1 ≤ x ≤ 3? What is the instantaneous rate of change of f(x) at the point (3, 7)? Explain. Explain why f(x) has a critical number in the interval 1 ≤ x ≤ 3? You can assume that f’(x) is continuous. In your explanation use the Mean Value Theorem, to argue that for some c, f’ (c) = 1. Then use the Intermediate Value Theorem applied to f’(x) to argue that for some d, f” (d) = 0.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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We are given that `f(1)=5,f(3)=7` and the tangent to the graph of `f(x)` at (3,7) is `-2x+13` . One possibility for `f(x)` is `f(x)=-3/2x^2+7x-1/2`

The graph of `f(x)` and the given tangent line:

** If f(x) is a quadratic, then `f(x)=ax^2+bx+c` . Using the two given points we have `f(1)=5==>a+b+c=5,f(3)=7==>9a+3b+c=7` . The derivative is `f'(x)=2ax+b` and `f'(3)=-2=2a(3)+b==>b=-2-6a`

Substituting for b we get the system `-5a+c=7,-9a+c=13` so `a=-3/2,b=7,c=-1/2` **

(1) The average rate of change of f(x)  on [1,3] is `(f(3)-f(1))/(3-1)=(7-5)/(3-1)=1`

(2) The instantaneous rate of change of f(x) at (3,7) is `f'(3)=-2` . This is the slope of the tangent line at x=3. For the example, `f'(x)=-3x+7==>f'(3)=-9+7=-2`

(3) How do we know that f(x) has a critical number on [1,3]? Assuming that f(x) and f'(x) are continuous on [1,3] we can apply the Mean Value Theorem: `f` continuous on `[a,b]` and differentiable on `(a,b)` implies that there exists a c in `(a,b)` such that `f'(c)=(f(b)-f(a))/(b-a)` . This there exists a `c in (1,3)` such that `f'(c)=(7-5)/(3-1)=1` . ` `

Now `f'(x)` is continuous by assumption on `[a,b]` , so we can apply the Intermediate Value Theorem to `f'(x)` . Then since  there is a c in (a,b) such that `f'(c)=1` and `f'(3)=-2` and `1>0> -2` then there is a `d in[a,b]` such that `f'(d)=0` .Since f'(x) is continuous, the point at x=d is a critical point. (The point in our example is x=7/3: `f'(7/3)=0` )

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