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Question: Find \( \vartriangle f \) and \( df \) for the function f for the indicated values of x, \( \vartria...

Find △f\vartriangle f and dfdf for the function f for the indicated values of x, △x\vartriangle x and compare f(x)=x3−2x2;x=2,△x=dx=0.5f(x) = {x^3} - 2{x^2};x = 2,\vartriangle x = dx = 0.5

Explanation

Solution

First, differentiation can be defined as the derivative of independent variables value and can be used to calculate feature independent variable per unit modification.
Let y=f(x)y = f(x) be the given function of x, the differentiation gives dydx\dfrac{{dy}}{{dx}} (with y-respect to)
The most popular power rule for differentiation is ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}

Complete step by step answer:
From the given that we have, f(x)=x3−2x2;x=2,△x=dx=0.5f(x) = {x^3} - 2{x^2};x = 2,\vartriangle x = dx = 0.5
Since f(x) is the function of y and both are equally as in the domain and codomain for x is the domain and y are the codomains.
Hence take f(x)=yf(x) = y , rewrite the given problem we get, y=x3−2x2;x=2,△x=dx=0.5y = {x^3} - 2{x^2};x = 2,\vartriangle x = dx = 0.5
By the differentiation rule, now we are going to differentiate the function y.
Thus, we get, y=x3−2x2⇒dydx=3x2−4xy = {x^3} - 2{x^2} \Rightarrow \dfrac{{dy}}{{dx}} = 3{x^2} - 4x
Now equating the denominator dx into the right-hand side values we get, dydx=3x2−4x⇒dy=(3x2−4x)dx\dfrac{{dy}}{{dx}} = 3{x^2} - 4x \Rightarrow dy = (3{x^2} - 4x)dx
From this we have some values for the function that x=2,△x=dx=0.5x = 2,\vartriangle x = dx = 0.5 where the value x is given as two and differentiation is given as zero points five.
subsisting the values in the converted equation we get, dy=(3x2−4x)dx⇒dy=(3(2)2−4(2))[0.5]dy = (3{x^2} - 4x)dx \Rightarrow dy = (3{(2)^2} - 4(2))[0.5]
Where x=2,△x=dx=0.5x = 2,\vartriangle x = dx = 0.5 .
Further solving the equation, we get, dy=(3(2)2−4(2))[0.5]⇒dy=(12−8)[0.5]⇒2dy = (3{(2)^2} - 4(2))[0.5] \Rightarrow dy = (12 - 8)[0.5] \Rightarrow 2 where 4×0.5=24 \times 0.5 = 2 .
Hence, we get the differentiation value of df=2df = 2
Now we are going to find the del value of the same function given Which is f(x)=x3−2x2f(x) = {x^3} - 2{x^2}
Now converting the function into del function thus we get, △f=f(x+△x)−f(x)\vartriangle f = f(x + \vartriangle x) - f(x)
Since from the given that we have x=2,△x=dx=0.5x = 2,\vartriangle x = dx = 0.5 , applying this we get, △f=f(2+0.5)−f(2)\vartriangle f = f(2 + 0.5) - f(2)
First, we will find the term one in the del value, f(2+0.5)f(2 + 0.5) , now convert this value into the f(x)=x3−2x2f(x) = {x^3} - 2{x^2} original function we get, f(2+0.5)=f(2.5)⇒(2.5)3−2(2.5)2f(2 + 0.5) = f(2.5) \Rightarrow {(2.5)^3} - 2{(2.5)^2} and further solving this we get, f(2.5)=3.125f(2.5) = 3.125
For the second term, we have f(2)=(2)3−2(2)2⇒0f(2) = {(2)^3} - 2{(2)^2} \Rightarrow 0
Hence, we get, △f=f(2+0.5)−f(2)⇒3.125\vartriangle f = f(2 + 0.5) - f(2) \Rightarrow 3.125
Therefore df=2df = 2 and △f=3.125\vartriangle f = 3.125 are the values of the given equation.

Note: Since the formation of the del x is the derivation of the given function represented as △f=f(x+△x)−f(x)\vartriangle f = f(x + \vartriangle x) - f(x) .
Derivative of any constant multiplied with the function f: ddx(a.y)=ay1\dfrac{d}{{dx}}(a.y) = a{y^1} (the constant values in the differentiation of integration will be not changed in any format)
The chain rule of the two functions is representing as dydx=dydu×dudx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}