Use the intermediate value theorem to show that f(x)=2x^4-x^3-4x^2-6 has a zero in the interval [-1.6,-1.5].

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The function f(x) = 2x^4 - x^3 - 4x^2 - 6 is a continuous function as `lim_(x->a) f(x) = f(a)` .

The value of f(-1.6) = 2(-1.6)^4-(-1.6)^3-4*(-1.6)^2-6 = 0.9632

f(-1.5) = 2(-1.5)^4-(-1.5)^3-4*(-1.5)^2-6 = -1.5

According to the intermediate value theorem a continuous function f(x) will take on a value f(x) = M that lies between f(a) and f(b) where f(a) < M < f(b) at at a value of x, such that a < x < b .

Taking M = 0, it can be seen that f(1.6)< M < f(-1.5).

This shows that f(x) = 0 or a root of f(x) lies in the interval of x given by [ -1.6, -1.5]

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