Question:

Find the approximate value of f (5.001), where f (x) = x3 − 7x2 + 15

Updated On: Sep 15, 2023
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Solution and Explanation

Let x = 5 and ∆x = 0.001. Then, we have:

f(5.001)=f(x+∆x)=(x+∆x)=(x+∆x)3-7(x+∆x)2+15

Now,∆y=f(x+∆x)-f(x)

≈f(x)+f'(x).∆x (asdx≈∆x)

f(5.001)≈(x3-7x2+15)+(3x2-14x)∆x

-35+(5)(0.001)

-31+0.005

-34.995

Hence, the approximate value of f (5.001) is −34.995

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Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
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