Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

{:[f^(')(x)=4e^(x)" and "],[f(2)=16+4e^(2).],[f(0)=◻]:}

f′(x)=4ex and f(2)=16+4e2.f(0)=□ \begin{array}{l}f^{\prime}(x)=4 e^{x} \text { and } f(2)=16+4 e^{2} . \\ f(0)=\square\end{array}

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Compare linear, exponential, and quadratic growthright chevron icon

Full solution

Q. f′(x)=4ex and f(2)=16+4e2.f(0)=□ \begin{array}{l}f^{\prime}(x)=4 e^{x} \text { and } f(2)=16+4 e^{2} . \\ f(0)=\square\end{array}
  1. Understand Given Information: Understand the given information.\newlineWe are given the derivative of a function f(x)f(x), which is f′(x)=4exf'(x) = 4e^x. We are also given the value of the function at x=2x = 2, which is f(2)=16+4e2f(2) = 16 + 4e^2. We need to find the value of the function at x=0x = 0, which is f(0)f(0).
  2. Integrate to Find f(x)f(x): Integrate the derivative to find the general form of f(x)f(x). Since f′(x)=4exf'(x) = 4e^x, we integrate to find f(x)f(x): f(x)=∫4exdx=4ex+Cf(x) = \int 4e^x dx = 4e^x + C, where CC is the constant of integration.
  3. Find Constant C: Use the given value f(2)=16+4e2f(2) = 16 + 4e^2 to find the constant C.\newlineWe plug x=2x = 2 into the general form of f(x)f(x):\newline16+4e2=4e2+C16 + 4e^2 = 4e^2 + C\newlineNow, solve for C:\newlineC=16+4e2−4e2C = 16 + 4e^2 - 4e^2\newlineC=16C = 16
  4. Write Specific Form: Write the specific form of f(x)f(x) with the found constant CC. Now that we have found CC, we can write the specific form of f(x)f(x): f(x)=4ex+16f(x) = 4e^x + 16
  5. Calculate f(0)f(0): Calculate f(0)f(0) using the specific form of f(x)f(x).\newlineWe plug x=0x = 0 into the specific form of f(x)f(x):\newlinef(0)=4e0+16f(0) = 4e^0 + 16\newlineSince e0=1e^0 = 1, we have:\newlinef(0)=4(1)+16f(0) = 4(1) + 16\newlinef(0)=4+16f(0) = 4 + 16\newlinef(0)=20f(0) = 20

More problems from Compare linear, exponential, and quadratic growth

Question
g(x)=2x+9g(□)=15 \begin{array}{l}g(x)=2 x+9 \\ g(\square)=15\end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
We want to factor the following expression:\newlinex4+10x+25 x^{4}+10 x+25 \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newline(x−2)4−(y+3)4 (x-2)^{4}-(y+3)^{4} \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newline(x−1)2+(x−1)+4 (x-1)^{2}+(x-1)+4 \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newline9x4+4y2 9 x^{4}+4 y^{2} \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newline16x4−25 16 x^{4}-25 \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newlinex4+3x2y+9y2 x^{4}+3 x^{2} y+9 y^{2} \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newline(x+5)4+2(x+5)2−1 (x+5)^{4}+2(x+5)^{2}-1 \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newline(x−3)2−10(x−3)+25 (x-3)^{2}-10(x-3)+25 \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
We want to factor the following expression:\newline(x+7)2+2y(x+7)+y2 (x+7)^{2}+2 y(x+7)+y^{2} \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

View all (29)