Consider the function f(x)=4x3+2x2+5, and let c be a number in the interval [01]. For what values of k is there a c in this interval such that ?
The Intermediate Value Theorem guarantees that there is a value c such that for 0≤c≤1 and f(0)≤k≤f(1), if f(1)≥f(0), and f(0)≥k≥f(1) if f(1)<f(0)."
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show that f satisfies the hypotheses of Rolle's theorem on [a,b],and find all number c in (a,b) such that f '(c)=0. f(x)=cos2x+2cosx ; [0,2(180)]
To find the values of k for which there exists a c in the interval [0,1] such that , we need to consider the function f(x) = 4x^3 + 2x^2 + 5.
Step 1: Set up the equation
We want to find c such that . In other words, we need to solve the equation 4c^3 + 2c^2 + 5 = k.
Step 2: Solve for c
Unfortunately, there is no general algebraic method to solve cubic equations like this one. However, we can use numerical methods or approximation techniques.
One approach is to use a graphing calculator or a software program to plot the graph of the function f(x) = 4x^3 + 2x^2 + 5, and then locate the points on the graph that correspond to the value k. This will give us an approximate value for c.
Step 3: Interval analysis
Since we are looking for a value of c in the interval [0,1], we can narrow down our search. We can evaluate f(0) and f(1) to get an idea of the behavior of the function within this interval.
When we substitute 0 into f(x), we get f(0) = 5.
When we substitute 1 into f(x), we get f(1) = 4 + 2 + 5 = 11.
This means that the function f(x) starts at 5 at x = 0 and increases towards 11 as x approaches 1.
Step 4: Answer
Based on our interval analysis and the graph of the function, we can conclude that there exists a value of c in the interval [0,1] such that for all k in the interval [5, 11].