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Question: If we have a function as \(f(x) = m{x^2} + nx + p\) then \({f^1}(1) + {f^1}(4) - {f^1}(5)\) is equal...

If we have a function as f(x)=mx2+nx+pf(x) = m{x^2} + nx + p then f1(1)+f1(4)−f1(5){f^1}(1) + {f^1}(4) - {f^1}(5) is equal to
1)m1)m
2)−m2) - m
3)n3)n
4)−n4) - n

Explanation

Solution

Differentiation is the process of derivation for the given functions or terms. Which is also the inverse of integration.
Differentiation can be defined as the derivative of the independent variable value and can be used to calculate features in an independence variance per unit modification.
Hence, we need to act on the first derivation of the given term f(x)=mx2+nx+pf(x) = m{x^2} + nx + p then we will substitute the values like x=1,4,5x = 1,4,5 which will lead us to the required answer.

Complete step-by-step solution:
Since from the given that we have the function f(x)=mx2+nx+pf(x) = m{x^2} + nx + p which is at the second-degree polynomial.
Now we need to act on the first derivation, due to the small difference in the process of derivation, we call it a differentiation function. So, the differentiation formula is dydx\dfrac{{dy}}{{dx}} . It shows that the difference in y is divided by the difference in x and also d is not the variable. Which can be represented in the polynomial as f1(x){f^1}(x) (first derivation of the polynomial)
Hence, we get the first derivation polynomial as f(x)=mx2+nx+p⇒f1(x)=2mx+nf(x) = m{x^2} + nx + p \Rightarrow {f^1}(x) = 2mx + n
Since in differentiation, the derivative of xx raised to the power is denoted by ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} .
Now substitute the values of x=1,4,5x = 1,4,5 then we see that f1(x)=2mx+n⇒f1(1)=2m+n{f^1}(x) = 2mx + n \Rightarrow {f^1}(1) = 2m + n where x=1x = 1 and f1(x)=2mx+n⇒f1(4)=8m+n{f^1}(x) = 2mx + n \Rightarrow {f^1}(4) = 8m + n where x=4x = 4 and f1(x)=2mx+n⇒f1(5)=10m+n{f^1}(x) = 2mx + n \Rightarrow {f^1}(5) = 10m + n where x=5x = 5
Thus, the required value is f1(1)+f1(4)−f1(5){f^1}(1) + {f^1}(4) - {f^1}(5) substituting the values we get f1(1)+f1(4)−f1(5)=2m+n+8m+n−10m−n{f^1}(1) + {f^1}(4) - {f^1}(5) = 2m + n + 8m + n - 10m - n
Further solving with the addition and subtraction operation we have f1(1)+f1(4)−f1(5)=2m+n+8m+n−10m−n=n{f^1}(1) + {f^1}(4) - {f^1}(5) = 2m + n + 8m + n - 10m - n = n
Therefore, the option 3)n3)n is correct.

Note: Differentiation and integration are inverse processes like a derivative of d(x2)dx=2x\dfrac{{d({x^2})}}{{dx}} = 2xand the integration is ∫2xdx=2x22⇒x2\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}
In differentiation, the derivative of xx raised to the power is denoted by ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} .
Integration is the process of inverse differentiation.
Process of finding the functions whose derivative is given named anti-differentiation or integration.
Integration is the process of adding the slices to find their whole.
It can be used to find the area, volume, and central points.
The integration is ∫abf(x)dx\int\limits_a^b {f(x)dx} the value of the anti-differentiation at the upper limit b and the lower limit a with the same anti-differentiation.