Given f ''(x) = 6x - 4 and f '(0)=1 and f(0)=5 find f '(x) and f (1)

Question:

Given {eq}f ''(x) = 6x - 4, \; f '(0)=1, \; \text{and} \; f(0)=5, \; \text{find} \; f '(x) \; \text{and} \; f (1) {/eq}

Using Integration to Find a Function


We are given the second derivative of a function as well as the values of the first derivative of the function and the function at fixed points. Using integration twice, we find the exact form of the function as well as the first derivative of the function and the function value at a different point on its domain.


Answer and Explanation:


Given {eq}f''(x)=6x-4 {/eq} we integrate both sides with respect to x to obtain

{eq}f'(x)=3x^2-4x+C \qquad (1) {/eq}

Using f'(0) = 1 in (1) gives us

{eq}f'(0)=1=3(0^2)-4(0)+C \implies C = 1. {/eq} which yields

{eq}f'(x)=3x^2-4x+1 \qquad (2) {/eq}

Next integrating (2) with respect to x as well leads to

{eq}f(x)=x^3-2x^2+x+D \qquad (3) {/eq}

Using f(0) = 5 in (3) then leads to

{eq}f(0)=5=0^3-2(0^2)+0+D \implies D = 5. {/eq}

So from (3)

{eq}f(x)=x^3-2x^2+x+5 \qquad (4) {/eq}


From (2), {eq}f'(x)=3x^2-4x+1 {/eq} and from (4), {eq}f(1)=1^3-2(1^2)+1+5 = 5. {/eq}


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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